Free math worksheets

- A linear approximation uses the known function value and derivative at the center point. - Measure how far the target input is from the center. - Apply the local linear change to the known function value.

1. The linear approximation at \(x=2\) is \(L(x)=f(2)+f'(2)(x-2)\). 2. Thus, \(L(x)=7-3(x-2)\). 3. Evaluate at \(x=2.04\): \(L(2.04)=7-3(0.04)=6.88\). 4. Therefore, \(f(2.04)\approx6.88\).

\(f(2.04)\approx6.88\)

- The derivative gives the tangent slope. - Parallel lines have equal slopes. - Set the derivative equal to \(1.5\). - Use the unit circle to solve the cosine equation.

1. Differentiate: \(f'(x)=3\cos x\). Thus, \(f'\left(\frac{\pi}{3}\right)=3\cdot\frac{1}{2}=1.5\). 2. A parallel tangent must have slope \(1.5\), so \(3\cos x=1.5\). Therefore, \(\cos x=\frac{1}{2}\). In \([0, 2\pi]\), the solutions are \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\).

1. The slope is \(1.5\). 2. \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\)

- A constant added outside a function creates a vertical shift. - Differentiate the logarithm and the added constant separately. - For the tangent line, combine the function value and derivative value at \(x=1\).

1. Adding \(3\) shifts the graph of \(y=\ln x\) upward by \(3\) units. 2. The derivative of \(\ln x\) is \(\frac1x\), and the constant has derivative \(0\), so \(f'(x)=\frac1x\). 3. At \(x=1\), \(f(1)=3\) and \(f'(1)=1\). The tangent line is \(y-3=x-1\), or \(y=x+2\).

a) Shift the graph of \(y=\ln x\) up \(3\) units. b) \(f'(x)=\frac1x\) c) \(y=x+2\)

- Use the product rule. - Recall the derivative of \(3^x\). - Evaluate the derivative at \(x=0\). - Use the origin and the slope to write the tangent line.

1. Use the product rule with \(u(x)=x\) and \(v(x)=3^x\): \(g'(x)=3^x+x(\ln 3)3^x\) \(=3^x(1+x\ln 3)\). 2. Since \(g(0)=0\), the graph passes through the origin. The tangent slope is \(g'(0)=3^0(1+0\ln 3)=1\). Therefore, the tangent line is \(y=x\).

a) \(g'(x)=3^x(1+x\ln 3)\) b) \(y=x\)

- Choose the nearby input where the cube root is easy to evaluate exactly. - Build the tangent-line approximation from the function value and derivative at that input. - Substitute the target input into the linearization, not into a new derivative calculation.

1. \(f(8)=2\), and \(f'(x)=\frac{1}{3x^{2/3}}\), so \(f'(8)=\frac{1}{12}\). 2. The linearization is \(L(x)=2+\frac{1}{12}(x-8)\). 3. Therefore, \(\sqrt[3]{8.3}\approx L(8.3)=2+\frac{0.3}{12}=2.025\).

\(\sqrt[3]{8.3}\approx2.025\)

- Treat volume as a function of the cube's side length. - A small input change can be estimated using the derivative at the original measurement. - The requested change is a volume, so check the units of the estimate.

1. The volume as a function of side length is \(V(s)=s^3\), so \(V'(s)=3s^2\). 2. Near \(s=10\), a small change \(\Delta s\) produces \(\Delta V\approx V'(10)\Delta s\). 3. With \(\Delta s=0.02\), \(\Delta V\approx3(10)^2(0.02)=6\,\text{cm}^3\).

The volume increases by approximately \(6\,\text{cm}^3\).

- Evaluate both the function and its derivative at the center \(a=0\). - Use \(L(x)=f(a)+f'(a)(x-a)\). - Substitute the nearby input into the linearization, not back into the original exponential function.

1. \(f(0)=1\) and \(f'(x)=e^x\), so \(f'(0)=1\). 2. Therefore, \(L(x)=f(0)+f'(0)(x-0)=1+x\). 3. \(e^{0.04}\approx L(0.04)=1.04\).

a) \(L(x)=1+x\) b) \(e^{0.04}\approx1.04\)

- The derivative gives the tangent slope. - Parallel lines have equal slopes; perpendicular nonvertical lines have slopes whose product is \(-1\). - Use unit-circle values and periodicity. - Substitute each x-coordinate into the original function.

1. The derivative is \(f'(x)=-\sin x\). 2. In part a, a parallel tangent must have slope \(-\frac{1}{2}\). Thus, \(-\sin x=-\frac{1}{2}\), so \(\sin x=\frac{1}{2}\). The solutions are \(x=\frac{\pi}{6}+2k\pi\) or \(x=\frac{5\pi}{6}+2k\pi\), where \(k\in\mathbb{Z}\). The corresponding y-coordinates are \(\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\). 3. In part b, a line perpendicular to \(y=x\) has slope \(-1\). Thus, \(-\sin x=-1\), so \(\sin x=1\). Therefore, \(x=\frac{\pi}{2}+2k\pi\), and the y-coordinate is \(0\).

a) \(\left(\frac{\pi}{6}+2k\pi, \frac{\sqrt{3}}{2}\right)\) and \(\left(\frac{5\pi}{6}+2k\pi, -\frac{\sqrt{3}}{2}\right)\), where \(k\in\mathbb{Z}\) b) \(\left(\frac{\pi}{2}+2k\pi, 0\right)\), where \(k\in\mathbb{Z}\)

- Differentiate using the constant multiple rule. - Set the derivative equal to the given slope. - Use the unit circle and periodicity to solve the trigonometric equation. - Substitute the inputs into the original function for the y-coordinates.

1. Differentiate: \(g'(x)=2\cos x\). 2. Set the derivative equal to the required slope: \(2\cos x=\sqrt{2}\), so \(\cos x=\frac{\sqrt{2}}{2}\). Therefore, \(x=\frac{\pi}{4}+2k\pi\) or \(x=-\frac{\pi}{4}+2k\pi\), where \(k\in\mathbb{Z}\). 3. Evaluate \(g\): \(g\left(\frac{\pi}{4}\right)=\sqrt{2}\) and \(g\left(-\frac{\pi}{4}\right)=-\sqrt{2}\).

\(\left(\frac{\pi}{4}+2k\pi, \sqrt{2}\right)\) and \(\left(-\frac{\pi}{4}+2k\pi, -\sqrt{2}\right)\), where \(k\in\mathbb{Z}\)

- The derivative gives the tangent slope. - Use the exact cosine value at \(\frac{2\pi}{3}\). - Write the line in point-slope form. - Keep all angle measures in radians.

1. Differentiate: \(f'(x)=\cos x\). 2. The tangent slope is \(m=f'\left(\frac{2\pi}{3}\right)=\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}\). 3. Use point-slope form: \(y-\frac{\sqrt{3}}{2}=-\frac{1}{2}\left(x-\frac{2\pi}{3}\right)\). Simplifying gives \(y=-\frac{1}{2}x+\frac{\pi}{3}+\frac{\sqrt{3}}{2}\).

\(y=-\frac{1}{2}x+\frac{\pi}{3}+\frac{\sqrt{3}}{2}\)

- Evaluate the function to find the point of tangency. - Differentiate using the constant multiple rule. - Use the exact sine and cosine values at \(\frac{\pi}{6}\). - Write the tangent line in point-slope form.

1. Find the point of tangency: \(g\left(\frac{\pi}{6}\right)=2\cos\left(\frac{\pi}{6}\right)=\sqrt{3}\). 2. Differentiate: \(g'(x)=-2\sin x\). The tangent slope is \(g'\left(\frac{\pi}{6}\right)=-2\cdot\frac{1}{2}=-1\). 3. Use point-slope form: \(y-\sqrt{3}=-\left(x-\frac{\pi}{6}\right)\). Therefore, \(y=-x+\frac{\pi}{6}+\sqrt{3}\).

\(y=-x+\frac{\pi}{6}+\sqrt{3}\)

- Evaluate the function and derivative at the given input. - Use point-slope form for the tangent line. - The derivative value is the tangent slope. - Choose the angle of inclination in the interval \([0^\circ, 180^\circ)\).

1. Evaluate the function: \(f\left(\frac{\pi}{2}\right)=0-2=-2\). Thus, the point is \(\left(\frac{\pi}{2}, -2\right)\). 2. Differentiate: \(f'(x)=-\sin x-2\cos x\). The tangent slope is \(f'\left(\frac{\pi}{2}\right)=-1\). 3. Use point-slope form: \(y+2=-\left(x-\frac{\pi}{2}\right)\). Therefore, \(y=-x+\frac{\pi}{2}-2\). 4. The angle of inclination is measured counterclockwise from the positive x-axis and satisfies \(\tan\alpha=-1\). Thus, \(\alpha=135^\circ\).

Tangent line: \(y=-x+\frac{\pi}{2}-2\) Angle of inclination: \(135^\circ\)

- Use the chain rule when differentiating \(\sin(2x)\). - Evaluate both the function and derivative at the given input. - Use point-slope form for the tangent line. - For a negative slope, choose the inclination angle in the second quadrant.

1. Find the point of tangency: \(f\left(\frac{\pi}{3}\right)=2\sin\left(\frac{2\pi}{3}\right)+\cos\left(\frac{\pi}{3}\right)=\sqrt{3}+\frac{1}{2}\). 2. Differentiate using the chain rule: \(f'(x)=4\cos(2x)-\sin x\). The tangent slope is \(m=f'\left(\frac{\pi}{3}\right)=-2-\frac{\sqrt{3}}{2}\). 3. Use point-slope form: \(y-\left(\sqrt{3}+\frac{1}{2}\right)=\left(-2-\frac{\sqrt{3}}{2}\right)\left(x-\frac{\pi}{3}\right)\). Equivalently, \(y=\left(-2-\frac{\sqrt{3}}{2}\right)x+\frac{2\pi}{3}+\frac{\pi\sqrt{3}}{6}+\sqrt{3}+\frac{1}{2}\). 4. The angle of inclination \(\alpha\in[0^\circ, 180^\circ)\) satisfies \(\tan\alpha=m\). Since the slope is negative, \(\alpha\approx 109.23^\circ\).

Tangent line: \(y-\left(\sqrt{3}+\frac{1}{2}\right)=\left(-2-\frac{\sqrt{3}}{2}\right)\left(x-\frac{\pi}{3}\right)\) Angle of inclination: \(\alpha\approx 109.23^\circ\)

- Set the two function values equal to find the intersection. - Use the sine and cosine derivative formulas for the tangent slopes. - Convert each slope to its inclination angle in \([0^\circ,180^\circ)\). - Use the smaller difference of the inclination angles and the requested precision.

1. At an intersection, \(2\sin x=2\cos x\). On \([0,\pi]\), this gives \(x=\frac{\pi}{4}\), with y-coordinate \(\sqrt{2}\). Thus, \(S=\left(\frac{\pi}{4},\sqrt{2}\right)\). 2. The derivatives are \(f'(x)=2\cos x\) and \(g'(x)=-2\sin x\). At \(S\), the slopes are \(\sqrt{2}\) and \(-\sqrt{2}\). 3. Therefore, \(\alpha=\arctan(\sqrt{2})\approx54.74^\circ\) and \(\beta=180^\circ-54.74^\circ\approx125.26^\circ\). 4. The smaller intersection angle is \(\gamma=\beta-\alpha\approx70.53^\circ\).

a) \(S=\left(\frac{\pi}{4},\sqrt{2}\right)\) b) \(\alpha\approx54.74^\circ\); \(\beta\approx125.26^\circ\) c) \(\gamma\approx70.53^\circ\)

- Evaluate the function and derivative at the given input. - Use point-slope form. - Relate the derivative value to the tangent slope. - Give the inclination angle in \([0^\circ, 180^\circ)\).

1. The point of tangency is \(\left(\frac{\pi}{2}, 1\right)\). 2. Differentiate: \(f'(x)=\cos x-\sin x\). The slope at \(x=\frac{\pi}{2}\) is \(-1\). 3. Use point-slope form: \(y-1=-\left(x-\frac{\pi}{2}\right)\). Therefore, \(y=-x+\frac{\pi}{2}+1\). 4. Since the slope is \(-1\), the angle of inclination is \(135^\circ\).

Tangent line: \(y=-x+\frac{\pi}{2}+1\) Angle of inclination: \(135^\circ\)

- Evaluate the function to find the point. - Differentiate cosine and keep the constant factor. - Use point-slope form for the tangent line. - A slope of \(-\sqrt{3}\) has an inclination angle in the second quadrant.

1. Evaluate the function: \(g\left(\frac{2\pi}{3}\right)=2\cdot\left(-\frac{1}{2}\right)+1=0\). Thus, the point is \(\left(\frac{2\pi}{3}, 0\right)\). 2. Differentiate: \(g'(x)=-2\sin x\). The tangent slope is \(g'\left(\frac{2\pi}{3}\right)=-\sqrt{3}\). 3. Use point-slope form: \(y=-\sqrt{3}\left(x-\frac{2\pi}{3}\right)\). Therefore, \(y=-\sqrt{3}x+\frac{2\pi\sqrt{3}}{3}\). 4. The standard angle of inclination in \([0^\circ, 180^\circ)\) is \(120^\circ\).

Tangent line: \(y=-\sqrt{3}x+\frac{2\pi\sqrt{3}}{3}\) Angle of inclination: \(120^\circ\)

- Evaluate the function at the given input. - Use the derivative to find the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Write both lines in point-slope form.

1. Find the point: \(f(\pi)=2\cos\pi+\pi=\pi-2\). Thus, \(P=(\pi, \pi-2)\). 2. Differentiate: \(f'(x)=-2\sin x+1\). The tangent slope is \(f'(\pi)=1\). Therefore, the tangent line is \(y-(\pi-2)=x-\pi\), or \(y=x-2\). 3. The normal slope is \(-1\). Thus, \(y-(\pi-2)=-(x-\pi)\), or \(y=-x+2\pi-2\).

Tangent line: \(y=x-2\) Normal line: \(y=-x+2\pi-2\)

- Evaluate the function to locate the point. - Differentiate sine and cosine carefully. - Use the derivative value as the tangent slope. - Use the negative reciprocal for the normal slope.

1. Evaluate the function: \(f\left(\frac{\pi}{2}\right)=1\). Thus, the point is \(\left(\frac{\pi}{2}, 1\right)\). 2. Differentiate: \(f'(x)=\cos x+\sin x\). The tangent slope is \(f'\left(\frac{\pi}{2}\right)=1\). Therefore, the tangent line is \(y-1=x-\frac{\pi}{2}\), or \(y=x-\frac{\pi}{2}+1\). 3. The normal slope is \(-1\). Thus, \(y-1=-\left(x-\frac{\pi}{2}\right)\), or \(y=-x+\frac{\pi}{2}+1\).

Tangent line: \(y=x-\frac{\pi}{2}+1\) Normal line: \(y=-x+\frac{\pi}{2}+1\)

- Evaluate the function at the given input. - Use the product rule to find the derivative. - Simplify with a double-angle identity if helpful. - Use negative reciprocal slopes for tangent and normal lines.

1. Evaluate the function: \(f\left(\frac{\pi}{2}\right)=1\). Thus, the point is \(\left(\frac{\pi}{2}, 1\right)\). 2. Differentiate using the product rule: \(f'(x)=\cos x(1+\cos x)-\sin^2x\). Using \(\cos^2x-\sin^2x=\cos(2x)\), \(f'(x)=\cos x+\cos(2x)\). Therefore, \(f'\left(\frac{\pi}{2}\right)=-1\). 3. The tangent line is \(y-1=-\left(x-\frac{\pi}{2}\right)\), or \(y=-x+\frac{\pi}{2}+1\). 4. The normal slope is \(1\), so the normal line is \(y-1=x-\frac{\pi}{2}\), or \(y=x-\frac{\pi}{2}+1\).

Tangent line: \(y=-x+\frac{\pi}{2}+1\) Normal line: \(y=x-\frac{\pi}{2}+1\)

- Evaluate the function at \(\pi\). - Use the chain rule for \(\cos^2x\). - The derivative value is the tangent slope. - Use the negative reciprocal for the normal slope.

1. Evaluate the function: \(f(\pi)=\cos^2\pi-\sin\pi=1\). Thus, \(P=(\pi, 1)\). 2. Differentiate using the chain rule: \(f'(x)=2\cos x(-\sin x)-\cos x\) \(=-\sin(2x)-\cos x\). Therefore, \(f'(\pi)=1\). 3. The tangent line is \(y-1=x-\pi\), or \(y=x-\pi+1\). 4. The normal slope is \(-1\), so the normal line is \(y-1=-(x-\pi)\), or \(y=-x+\pi+1\).

Tangent line: \(y=x-\pi+1\) Normal line: \(y=-x+\pi+1\)

- Find both \(f(-1)\) and \(f'(-1)\). - Use point-slope form for the tangent line. - Keep \(e\) symbolic in the exact equation. - Use the inverse tangent of the slope to find the angle.

1. Find the point of tangency: \(f(-1)=10e^{-1}+1=\frac{10}{e}+1\). 2. Differentiate: \(f'(x)=10e^x+2x\). The slope at \(x=-1\) is \(m=f'(-1)=\frac{10}{e}-2\). 3. Use point-slope form: \(y-\left(\frac{10}{e}+1\right)=\left(\frac{10}{e}-2\right)(x+1)\). Simplifying gives \(y=\left(\frac{10}{e}-2\right)x+\frac{20}{e}-1\). 4. The angle of inclination satisfies \(\tan\theta=m\). Therefore, \(\theta=\arctan\left(\frac{10}{e}-2\right)\approx 59.2^\circ\).

Tangent line: \(y=\left(\frac{10}{e}-2\right)x+\frac{20}{e}-1\) Angle of inclination: \(\theta\approx 59.2^\circ\)

- Evaluate the function to find the point of tangency. - Differentiate using the chain rule. - Use the derivative value as the tangent slope. - The normal slope is the negative reciprocal of the tangent slope.

1. Find the point: \(f(0)=e^0+e^0=2\), so the point is \((0, 2)\). 2. Differentiate using the chain rule: \(f'(x)=2e^{2x}+e^x\). Thus, the tangent slope is \(m_t=f'(0)=2+1=3\). The tangent line is \(t:y-2=3(x-0)\), or \(t:y=3x+2\). 3. The normal slope is the negative reciprocal: \(m_n=-\frac{1}{3}\). Therefore, \(n:y-2=-\frac{1}{3}(x-0)\), or \(n:y=-\frac{1}{3}x+2\).

\(t:y=3x+2\) \(n:y=-\frac{1}{3}x+2\)

- Use the product rule to find the derivative. - Evaluate the function and derivative at \(x=0\). - Write the tangent line in point-slope form. - Use the negative reciprocal for the normal slope.

1. Find the point: \(f(0)=(-2)e^0=-2\), so the point is \((0, -2)\). 2. Differentiate using the product rule: \(f'(x)=e^x+(x-2)e^x=(x-1)e^x\). Thus, the tangent slope is \(m_t=f'(0)=-1\). The tangent line is \(y+2=-1(x-0)\), or \(y=-x-2\). 3. The normal slope is \(m_n=-\frac{1}{m_t}=1\). Therefore, the normal line is \(y+2=x\), or \(y=x-2\).

Tangent line: \(y=-x-2\) Normal line: \(y=x-2\)

- Use the linearization formula centered at \(h_0=20\). - A linear approximation replaces a nearby function value with the corresponding tangent-line value. - Find absolute error by taking the absolute value of the difference between the approximation and the exact model value.

1. At \(h=20\), \(v(20)=\sqrt{392.4}\approx19.809\). 2. Differentiate: \(v'(h)=\frac{9.81}{\sqrt{19.62h}}\), so \(v'(20)\approx0.495\). 3. The linearization is \(L(h)\approx19.809+0.495(h-20)\). 4. \(L(21)\approx19.809+0.495=20.304\,\text{m/s}\). 5. The exact model value is \(v(21)=\sqrt{412.02}\approx20.298\,\text{m/s}\). The absolute error is \(|20.304-20.298|\approx0.006\,\text{m/s}\).

a) \(L(h)\approx19.809+0.495(h-20)\) b) \(L(21)\approx20.304\,\text{m/s}\) c) \(v(21)\approx20.298\,\text{m/s}\); absolute error \(\approx0.006\,\text{m/s}\)

- Write the tangent line at a general input \(u\). - A line through the origin has y-intercept \(0\). - The tangent slope is the derivative value at the point of tangency. - Use logarithm properties for powers of \(e\).

1. Differentiate: \(f'(x)=\frac{1}{x}\). 2. At \(x_0=e^2\), \(f(e^2)=2\) and \(f'(e^2)=\frac{1}{e^2}\). Therefore, \(y=\frac{1}{e^2}(x-e^2)+2=\frac{x}{e^2}+1\). 3. The tangent line at a general input \(u>0\) is \(y=\frac{1}{u}(x-u)+\ln u=\frac{x}{u}+\ln u-1\). For this line to pass through the origin, \(\ln u-1=0\), so \(u=e\). The point of tangency is \((e, 1)\), and the tangent line is \(y=\frac{x}{e}\). 4. At \(x_1=\sqrt{e}\), the tangent line is \(y=\frac{x}{\sqrt{e}}-\frac{1}{2}\). Setting \(y=0\) gives the x-intercept \(x=\frac{\sqrt{e}}{2}\).

a) \(y=\frac{x}{e^2}+1\) b) Point of tangency: \((e, 1)\); tangent line: \(y=\frac{x}{e}\) c) x-intercept: \(x=\frac{\sqrt{e}}{2}\)

- Parallel lines have equal slopes. - Write the tangent line at a general input \(u\). - A point on the y-axis determines the tangent’s y-intercept. - Set \(x=0\) in the general tangent equation.

1. A line parallel to \(y=x\) has slope \(1\). Since \(f'(x)=\frac{1}{x}\), solve \(\frac{1}{x}=1\), giving \(x=1\). Since \(f(1)=0\), the tangent line is \(y=x-1\). 2. The tangent line at a general input \(u>0\) is \(y=\frac{x}{u}+\ln u-1\). Passing through \((0, -2)\) requires \(\ln u-1=-2\), so \(u=e^{-1}\). The slope is \(1/u=e\), giving \(y=ex-2\). 3. From the general tangent equation, setting \(x=0\) gives \(y=\ln u-1\), so the y-intercept is \((0, \ln u-1)\). For an intercept of \(3\), \(\ln u-1=3\), so \(u=e^4\).

a) \(y=x-1\) b) \(y=ex-2\) c) The y-intercept is \((0, \ln u-1)\), and \(u=e^4\) when the intercept is \(3\).

- Find where the squared expression equals zero. - Apply the chain rule carefully. - A tangent line requires a point and a slope. - Evaluate both the function and derivative at \(x=1\).

1. The logarithm’s argument is zero only when \(2x-4=0\), so \(x=2\). It is positive for every other real input. Therefore, \(D_k=\mathbb{R}\setminus\{2\}\). 2. Differentiate using the chain rule: \(k'(x)=\frac{4(2x-4)}{(2x-4)^2}=\frac{2}{x-2}\). At \(x=1\), \(k(1)=\ln4\) and \(k'(1)=-2\). Thus, the tangent line is \(y=-2(x-1)+\ln4=-2x+2+\ln4\).

a) \(D_k=\mathbb{R}\setminus\{2\}\) b) \(y=-2x+2+\ln4\)

- Use the fact that \(\ln1=0\). - Apply the chain rule to find the derivative. - A tangent line requires a point and a slope. - Use point-slope form.

1. To find the zero, solve \(\ln(4-x)=0\). Thus, \(4-x=1\), so \(x=3\). 2. Differentiate: \(f'(x)=-\frac{1}{4-x}=\frac{1}{x-4}\). At \(x=3\), \(f(3)=0\) and \(f'(3)=-1\). Therefore, the tangent line is \(y=-(x-3)=-x+3\).

Zero: \(x=3\) Tangent line: \(y=-x+3\)

- Read the effects of the factor and constant from the formula. - Parallel lines have equal slopes. - Set the derivative equal to the target slope. - Substitute the resulting input into the original function.

1. Multiply the y-values of \(\ln x\) by \(2\), reflect the graph across the x-axis, and then shift it up \(1\) unit. 2. The given line has slope \(-1\). Differentiate: \(g'(x)=-\frac{2}{x}\). Set the derivative equal to \(-1\): \(-\frac{2}{x}=-1\), so \(x=2\). The corresponding y-value is \(g(2)=1-2\ln2\).

a) Vertically stretch by a factor of \(2\), reflect across the x-axis, and shift up \(1\) unit. b) \((2, 1-2\ln2)\)

- Use the chain rule for \(f\). - A tangent line needs a point and a derivative value. - Equal exponential expressions with the same base have equal exponents. - Compare the two derivative values at the intersection input.

1. By the chain rule, \(f'(x)=2e^{2x-4}\). At \(x=2\), \(f(2)=1\) and \(f'(2)=2\). Thus, the tangent line is \(y-1=2(x-2)\), or \(y=2x-3\). 2. At an intersection, \(e^{2x-4}=e^x\). Since the bases are equal and the exponential function is one-to-one, \(2x-4=x\), so \(x=4\). 3. Since \(g'(x)=e^x\), \(g'(4)=e^4\). Also, \(f'(4)=2e^{2\cdot4-4}=2e^4\). Therefore, \(f'(4)=2g'(4)\).

a) \(f'(x)=2e^{2x-4}\); tangent line: \(y=2x-3\) b) The x-coordinate of \(S\) is \(4\). c) \(f'(4)=2e^4=2g'(4)\).

- Build \(L(x)\) from the function value and derivative at the center. - Match each logarithm to the form \(\ln(1+x)\). - Use the sign of the second derivative to decide where the graph lies relative to its tangent line.

1. Since \(f(0)=\ln(1)=0\) and \(f'(x)=\frac{1}{x+1}\), \(f'(0)=1\). 2. The linearization is \(L(x)=f(0)+f'(0)(x-0)=x\). Thus \(\ln(1+x)\approx x\) for \(x\) near \(0\). 3. For \(\ln(1.1)\), use \(x=0.1\): \(\ln(1.1)\approx L(0.1)=0.1\). 4. For \(\ln(1.02)\), use \(x=0.02\): \(\ln(1.02)\approx L(0.02)=0.02\). 5. For \(\ln(0.95)\), use \(x=-0.05\): \(\ln(0.95)\approx L(-0.05)=-0.05\). 6. The second derivative is \(f''(x)=-\frac{1}{(x+1)^2}<0\) for \(x>-1\). Therefore, \(f\) is concave down, and its tangent lines lie above the graph. Each approximation is greater than the corresponding exact value.

a) \(L(x)=x\) b) \(\ln(1.1)\approx0.1\); \(\ln(1.02)\approx0.02\); \(\ln(0.95)\approx-0.05\) c) The graph is concave down on \((-1,\infty)\), so all three linear approximations are overestimates.

- Use the point-slope form of the tangent line at \(x=4\). - Substitute each nearby x-value into your linear approximation. - Relate the sign of the second derivative to the position of a tangent line.

1. Since \(g(4)=2\) and \(g'(x)=\frac{1}{2\sqrt{x}}\), \(g'(4)=\frac{1}{4}\). 2. The tangent-line approximation is \(L(x)=2+\frac{1}{4}(x-4)=\frac{1}{4}x+1\). 3. \(L(4.1)=2+\frac{1}{4}(0.1)=2.025\). 4. \(L(4.04)=2+\frac{1}{4}(0.04)=2.01\). 5. \(L(3.9)=2+\frac{1}{4}(-0.1)=1.975\). 6. The second derivative is \(g''(x)=-\frac{1}{4x^{3/2}}<0\) for \(x>0\). Thus the graph is concave down, so its tangent line lies above the graph except at the point of tangency. Therefore, all three approximations are overestimates.

a) \(L(x)=\frac{1}{4}x+1\) b) \(\sqrt{4.1}\approx2.025\); \(\sqrt{4.04}\approx2.01\); \(\sqrt{3.9}\approx1.975\) c) Since \(g''(x)<0\) for \(x>0\), the graph is concave down and the tangent-line values are overestimates.

- Recall the derivative of \(\log_bx\). - Find the function value and derivative value at \(x=3\). - Use point-slope form. - Convert between base-3 and natural logarithms if needed.

1. The point on the graph is \(f(3)=\log_33=1\). 2. Differentiate: \(f'(x)=\frac{1}{x\ln3}\). Thus, the slope at \(x=3\) is \(f'(3)=\frac{1}{3\ln3}\). 3. Use point-slope form: \(y-1=\frac{1}{3\ln3}(x-3)\). Therefore, \(y=\frac{x}{3\ln3}+1-\frac{1}{\ln3}\).

\(y=\frac{x}{3\ln3}+1-\frac{1}{\ln3}\)

- Find the function value and derivative value at \(x=1\). - Use the tangent line as the linearization. - Substitute the nearby inputs into the linearization. - Absolute error is the absolute difference between the estimate and actual value.

1. Since \(f'(x)=\frac{1}{x}\), \(f(1)=0\) and \(f'(1)=1\). Therefore, the tangent line is \(L(x)=x-1\). 2. Use \(\ln x\approx L(x)\) near \(x=1\): \(\ln(1.08)\approx0.08\) and \(\ln(0.95)\approx-0.05\). 3. Since \(\ln(1.08)\approx0.076961\), the absolute error is \(|0.08-\ln(1.08)|\approx0.0030\).

1. \(L(x)=x-1\) 2. \(\ln(1.08)\approx0.08\); \(\ln(0.95)\approx-0.05\) 3. Absolute error: \(0.0030\)

- Find the function value and slope at \(x=0\). - Write each logarithm in the form \(\ln(1+x)\). - The tangent line gives a nearby approximation. - Consider what happens to \(\frac{1}{n}\) as \(n\) grows.

1. The function satisfies \(g(0)=0\) and \(g'(x)=\frac{1}{1+x}\), so \(g'(0)=1\). Therefore, the tangent line at \(x=0\) is \(y=x\), which gives \(\ln(1+x)\approx x\) near \(0\). 2. For \(1.003=1+0.003\), \(\ln(1.003)\approx0.003\). For \(0.998=1-0.002\), \(\ln(0.998)\approx-0.002\). 3. When \(n\) is large, \(\frac{1}{n}\) is small, so \(\ln\left(1+\frac{1}{n}\right)\approx\frac{1}{n}\).

1. The tangent line to \(g(x)=\ln(1+x)\) at \(x=0\) is \(y=x\). 2. \(\ln(1.003)\approx0.003\); \(\ln(0.998)\approx-0.002\) 3. \(\ln\left(1+\frac{1}{n}\right)\approx\frac{1}{n}\)

- Find the point on the graph at \(x=1\). - Apply the product rule. - Evaluate the derivative at \(x=1\). - Use point-slope form.

1. Evaluate the function: \(f(1)=1^2\ln1=0\). 2. Differentiate using the product rule: \(f'(x)=2x\ln x+x\). At \(x=1\), \(f'(1)=1\). 3. The tangent line through \((1, 0)\) with slope \(1\) is \(y=x-1\).

\(y=x-1\)

- Compare the three formulas term by term. - A tangent line requires both \(f(x_0)\) and \(f'(x_0)\). - A horizontal tangent has derivative value \(0\).

1. Since \(f(x)=g(x)+h(x)\), the y-coordinate on \(f\) is the sum of the corresponding y-coordinates on \(g\) and \(h\). At \(x=0\), \(f(0)=1\). 2. Differentiate: \(f'(x)=0.5-\sin x\). At \(x=\pi\), \(f'(\pi)=0.5\) and \(f(\pi)=\frac{\pi}{2}-1\). Thus, the tangent is \(t(x)=0.5x-1\). 3. Horizontal tangents satisfy \(\sin x=\frac{1}{2}\). On \([0,2\pi]\), the solutions are \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\).

a) Add corresponding y-values: \(f(x)=g(x)+h(x)\). At \(x=0\), \(f(0)=1\). b) \(t(x)=0.5x-1\) c) \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)

- Use the point and slope at the tangency input to build \(L\). - The target input is close to the point of tangency. - For the error direction, compare the vertical positions of the curve and tangent near the target input.

1. \(f(3)=2\), and \(f'(x)=\frac{1}{2\sqrt{x+1}}\), so \(f'(3)=\frac{1}{4}\). 2. Thus, \(L(x)=2+\frac{1}{4}(x-3)=\frac{1}{4}x+\frac{5}{4}\). 3. \(L(3.8)=2.2\), so \(f(3.8)\approx2.2\). 4. Near \(x=3.8\), the displayed tangent line lies above the curve. Therefore, \(2.2\) is an overestimate.

a) \(L(x)=\frac{1}{4}x+\frac{5}{4}\), so \(f(3.8)\approx2.2\). b) The approximation is an overestimate.

- Build each tangent line from the function value and derivative at its own center. - Compare how far the target input lies from each center. - Use the graph to check which tangent follows the logarithm more closely near the target.

1. At \(x=1\), \(f(1)=0\) and \(f'(1)=1\), so \(L_1(x)=x-1\). Hence, \(L_1(1.9)=0.9\). 2. At \(x=2\), \(f(2)=\ln2\) and \(f'(2)=\frac{1}{2}\), so \(L_2(x)=\ln2+\frac{1}{2}(x-2)\). Hence, \(L_2(1.9)=\ln2-0.05\approx0.6431\). 3. The input \(1.9\) is much closer to the center \(2\) than to the center \(1\), and the graph shows \(L_2\) tracking \(f\) closely near \(1.9\). Therefore, \(L_2\) should give the better approximation.

a) From \(L_1\): \(0.9\). From \(L_2\): \(\ln2-0.05\approx0.6431\). b) \(L_2\) should be better because \(1.9\) is closer to its center \(x=2\), and the graph shows closer local agreement there.

- Use the exact sine and cosine values at the convenient center angle. - The nearby input differs from the center by only \(0.02\) radians. - The sign of the second derivative determines whether the graph bends above or below its tangent line locally.

1. \(f\left(\frac{\pi}{6}\right)=\frac{1}{2}\) and \(f'(x)=\cos x\), so \(f'\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\). 2. Thus \(L(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)\). 3. At \(x=\frac{\pi}{6}+0.02\), \(L(x)=\frac{1}{2}+0.01\sqrt{3}\approx0.51732\). 4. Since \(f''(x)=-\sin x<0\) near \(x=\frac{\pi}{6}\), the graph is concave down there, so the tangent line lies above the graph. The linear approximation is an overestimate.

a) \(L(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)\) b) \(\sin\left(\frac{\pi}{6}+0.02\right)\approx\frac{1}{2}+0.01\sqrt{3}\approx0.51732\) c) Overestimate, because \(f''(x)=-\sin x<0\) near \(\frac{\pi}{6}\).

- Parallel lines have equal slopes. - Use the double-angle identity for sine. - Perpendicular nonvertical lines have a slope product of \(-1\). - Use the maximum possible values of sine and cosine in part d.

1. The derivatives are \(f'(x)=\cos x\) and \(g'(x)=-\sin x\). Parallel tangents have equal slopes, so \(\cos x=-\sin x\). On \([0, \pi]\), this gives \(x_P=\frac{3\pi}{4}\). 2. The product of the slopes is \(f'(x)g'(x)=-\sin x\cos x=-\frac{1}{2}\sin(2x)\). 3. Perpendicular nonvertical lines require a slope product of \(-1\). That would require \(-\frac{1}{2}\sin(2x)=-1\), or \(\sin(2x)=2\), which is impossible. Therefore, no common input produces perpendicular tangents. 4. For different inputs, require \(\cos(x_1)[-\sin(x_2)]=-1\), so \(\cos(x_1)\sin(x_2)=1\). On \([0, \pi]\), this occurs when \(x_1=0\) and \(x_2=\frac{\pi}{2}\).

a) \(x_P=\frac{3\pi}{4}\) b) \(f'(x)g'(x)=-\frac{1}{2}\sin(2x)\) c) No such common input exists. d) \(x_1=0\), \(x_2=\frac{\pi}{2}\)

- Tangent graphs have equal values and equal slopes at the same input. - Differentiate both functions. - Use the double-angle identity for sine. - Solve the derivative equation first, then use the value equation.

1. At a point of tangency, the function values and derivatives must be equal: \(f(x)=g_c(x)\) and \(f'(x)=g_c'(x)\). 2. The derivatives are \(f'(x)=2\cos x\) and \(g_c'(x)=-2\sin(2x)\). Set them equal and use \(\sin(2x)=2\sin x\cos x\): \(2\cos x=-4\sin x\cos x\), so \(2\cos x(1+2\sin x)=0\). 3. If \(\cos x=0\), then \(\sin x=1\) or \(-1\), and \(\cos(2x)=-1\). For \(\sin x=1\), the value equation gives \(2=-1+c\), so \(c=3\). For \(\sin x=-1\), it gives \(-2=-1+c\), so \(c=-1\). 4. If \(\sin x=-\frac{1}{2}\), then \(\cos(2x)=\frac{1}{2}\). The value equation gives \(-1=\frac{1}{2}+c\), so \(c=-\frac{3}{2}\).

\(c\in\left\{-\frac{3}{2},-1,3\right\}\)

- Tangency requires equal function values and equal derivatives. - Use the chain rule to differentiate \(\sin^2x\). - Factor the derivative equation. - Use \(\sin^2x+\cos^2x=1\) in the remaining case.

1. At a point of tangency, \(f(x)=h_k(x)\) and \(f'(x)=h_k'(x)\). 2. Differentiate: \(f'(x)=2\sin x\cos x\) and \(h_k'(x)=\sin x\). Set the slopes equal: \(\sin x(2\cos x-1)=0\). 3. If \(\sin x=0\), then \(\cos x=1\) or \(-1\). When \(\cos x=1\), the value equation gives \(0=k-1\), so \(k=1\). When \(\cos x=-1\), it gives \(0=k+1\), so \(k=-1\). 4. If \(\cos x=\frac{1}{2}\), then \(\sin^2x=1-\frac{1}{4}=\frac{3}{4}\). The value equation gives \(\frac{3}{4}=k-\frac{1}{2}\), so \(k=\frac{5}{4}\).

\(k\in\left\{-1,1,\frac{5}{4}\right\}\)

- Find the tangent line using the derivative at \(x=3\). - Write the absolute difference between the function and tangent line. - Solving a squared inequality produces an absolute-value condition. - The endpoints occur where the error equals the allowed maximum.

1. Since \(f(3)=2\) and \(f'(x)=2x-4\), the tangent slope is \(f'(3)=2\). Therefore, the tangent line is \(y-2=2(x-3)\), or \(y=2x-4\). 2. The absolute error is \(|f(x)-(2x-4)|=|x^2-6x+9|=(x-3)^2\). 3. Require \((x-3)^2\le 0.04\). Then \(|x-3|\le 0.2\). 4. Therefore, \(2.8\le x\le 3.2\), so the largest interval is \([2.8,3.2]\).

1. \(y=2x-4\) 2. \([2.8,3.2]\)

- Linearize the volume with respect to radius at the design radius. - For an error bound, use magnitudes of the small input and output changes. - In part b, work backward from the allowed output error to the permitted input error.

1. \(V'(r)=4\pi r^2\), so at \(r=10\), \(V'(10)=400\pi\). 2. For a small radius error \(\Delta r\), the volume error satisfies \(|\Delta V|\approx400\pi|\Delta r|\). 3. With \(|\Delta r|=0.03\), \(|\Delta V|\approx12\pi\,\text{cm}^3\approx37.70\,\text{cm}^3\). 4. For the quality standard, require \(400\pi|\Delta r|\le20\). Thus, \(|\Delta r|\le\frac{1}{20\pi}\,\text{cm}\approx0.0159\,\text{cm}\).

a) Approximately \(12\pi\,\text{cm}^3\approx37.70\,\text{cm}^3\). b) At most \(\frac{1}{20\pi}\,\text{cm}\approx0.0159\,\text{cm}\).