Free math worksheets

- Compare the expression with the h-form definition of a derivative at a point. - Identify the fixed input that appears in both function values. - The derivative notation should use that fixed input.

1. The h-form of the derivative at \(x=a\) is \(f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\). 2. Here \(a=4\), so the limit represents \(f'(4)\).

\(f'(4)\)

- Think about the geometric difference between a secant line and a tangent line. - Consider what happens when the two points defining a secant move together. - A derivative at one point is connected to a limiting slope.

1. As \(Q\) moves toward \(P\), the secant line approaches the tangent line to the graph at \(P\). 2. The limiting secant slope is the derivative at the x-coordinate of \(P\). 3. Since \(P=(1,1)\), that derivative quantity is \(f'(1)\).

The secant approaches the tangent line at \(P\), and its slope approaches \(f'(1)\).

- Start with the limit definition of the derivative. - Notice what happens to the constant terms in the numerator. - What shape is the graph of a linear function? - Does a line’s slope change from one point to another?

1. At \(x = -2\), \(\frac{f(-2+h)-f(-2)}{h} = \frac{-1.5(-2+h)+3-6}{h} = \frac{-1.5h}{h} = -1.5\) for \(h \ne 0\). Therefore, \(f'(-2) = \lim_{h\to 0}(-1.5) = -1.5\). 2. At \(x = 4\), \(\frac{f(4+h)-f(4)}{h} = \frac{-1.5(4+h)+3-(-3)}{h} = \frac{-1.5h}{h} = -1.5\) for \(h \ne 0\). Therefore, \(f'(4) = -1.5\). 3. The graph is a line with constant slope \(-1.5\). At every point, the tangent line is the same line, so its slope and the derivative are constant.

1. \(f'(-2) = -1.5\) and \(f'(4) = -1.5\) 2. Both equal the constant slope of the line. The tangent line has the same slope at every point.

- Distinguish between \(f(x)\) and \(f'(x)\). - Which words describe a function value, and which describe a derivative? - Pay attention to which number is the input and which is the output or slope. - A point on the graph has coordinates \((x, f(x))\).

1. Statements A, D, and F all say that the function value at \(x = 5\) is \(2\). The coordinate statement \((5, 2)\) has the form \((x, f(x))\). 2. Statements B, C, and E all say that the derivative at \(x = 2\) is \(5\). The derivative is both the tangent-line slope and the instantaneous rate of change. 3. Statement G concerns the derivative at \(x = 5\), and statement H concerns the function value at \(x = 2\), so neither belongs to the two groups.

Group 1: A, D, F Group 2: B, C, E Not in either group: G, H

- Compare \(f(x_0+h)\) with \(f(x_0)\). - Determine what happens to the numerator of the difference quotient. - Think about the slope of a horizontal line. - Relate that slope to the derivative.

1. Form the difference quotient: \(\frac{f(x_0+h)-f(x_0)}{h}\). 2. Because \(f(x) = -8\) for every input, \(\frac{f(x_0+h)-f(x_0)}{h} = \frac{-8-(-8)}{h} = 0\) for \(h\ne 0\). 3. Take the limit: \(f'(x_0) = \lim_{h\to 0}0 = 0\). 4. The graph is the horizontal line \(y = -8\), so its slope is \(0\) at every point. This agrees with the derivative.

1. \(f'(x_0) = 0\) for every \(x_0\in\mathbb{R}\). 2. The graph is horizontal, so its slope is \(0\) everywhere.

- The derivative of a position function represents instantaneous velocity. - Decide whether the car's position changes as time passes. - Substitute the constant position values into the difference quotient. - Interpret a zero rate of change in the motion context.

1. Form the difference quotient at \(t = 10\): \(\frac{s(10+h)-s(10)}{h} = \frac{15-15}{h} = 0\) for \(h\ne 0\). 2. Take the limit: \(s'(10) = \lim_{h\to 0}0 = 0\). 3. The derivative of position is instantaneous velocity. Therefore, the car's velocity at \(t = 10\) is \(0\,\frac{\text{m}}{\text{s}}\), which means the car is not moving.

1. \(s'(10) = 0\,\frac{\text{m}}{\text{s}}\) 2. The instantaneous velocity is \(0\,\frac{\text{m}}{\text{s}}\), so the car is stationary.

- Write the limit definition of the derivative at a specific point. - Substitute \(x = -2\) into the definition. - Simplify the numerator until you can factor out \(h\). - After canceling \(h\), evaluate the limit as \(h\) approaches \(0\).

1. Evaluate the function at the point: \(f(-2) = -0.5(-2)^2 + 4(-2) = -10\). 2. Set up the difference quotient: \(\frac{f(-2+h)-f(-2)}{h} = \frac{-0.5(-2+h)^2+4(-2+h)-(-10)}{h}\). 3. Simplify the numerator: \(-0.5(4-4h+h^2)-8+4h+10 = 6h-0.5h^2\). 4. For \(h \ne 0\), the quotient simplifies to \(\frac{6h-0.5h^2}{h} = 6-0.5h\). 5. Take the limit: \(f'(-2) = \lim_{h\to 0}(6-0.5h) = 6\).

\(f'(-2) = 6\), so the tangent line has slope \(6\).

- Instantaneous rate of change is the derivative at the given point. - Set up and simplify the difference quotient. - Use the square of a binomial carefully when expanding \((3+h)^2\). - Determine what remains as \(h\) approaches \(0\).

1. Evaluate \(f(3)\): \(f(3) = 2(3)^2-4(3)+1 = 7\). 2. Form the difference quotient: \(\frac{f(3+h)-f(3)}{h} = \frac{2(3+h)^2-4(3+h)+1-7}{h}\). 3. Expand and combine like terms in the numerator: \(2(9+6h+h^2)-12-4h-6 = 8h+2h^2\). 4. For \(h \ne 0\), simplify: \(\frac{8h+2h^2}{h} = 8+2h\). 5. Take the limit: \(f'(3) = \lim_{h\to 0}(8+2h) = 8\).

The instantaneous rate of change is \(f'(3) = 8\).

- First evaluate the function at the given point. - Substitute \(2+h\) into the function and expand carefully. - Simplify until \(h\) can be factored and canceled. - Evaluate the remaining expression as \(h\) approaches \(0\).

1. Evaluate \(f(2)\): \(f(2) = 2(2)^2-5(2)+1 = -1\). 2. Form the difference quotient: \(\frac{f(2+h)-f(2)}{h} = \frac{2(2+h)^2-5(2+h)+1-(-1)}{h}\). 3. Expand and simplify the numerator: \(2(4+4h+h^2)-10-5h+2 = 2h^2+3h\). 4. For \(h \ne 0\), simplify the quotient to \(2h+3\). 5. Take the limit: \(f'(2) = \lim_{h\to 0}(2h+3) = 3\).

\(f'(2) = 3\)

- Instantaneous rate of change is the derivative at the given point. - Expand the squared binomial carefully. - Remember to subtract \(g(4)\) in the numerator. - After canceling \(h\), take the limit as \(h\) approaches \(0\).

1. Evaluate \(g(4)\): \(g(4) = -0.5(4)^2+3(4) = 4\). 2. Form the difference quotient: \(\frac{g(4+h)-g(4)}{h} = \frac{-0.5(4+h)^2+3(4+h)-4}{h}\). 3. Expand and simplify the numerator: \(-0.5(16+8h+h^2)+12+3h-4 = -0.5h^2-h\). 4. For \(h \ne 0\), simplify the quotient to \(-0.5h-1\). 5. Take the limit: \(g'(4) = \lim_{h\to 0}(-0.5h-1) = -1\).

The instantaneous rate of change is \(g'(4) = -1\).

- Distinguish between \(f'(x)\) and \(f(x)\). - Recall the geometric meaning of a derivative. - Compare Statement 4 with the limit definition of the derivative. - Compare the tangent-line slope with the slope of the given line.

1. Statement 1 directly restates the notation \(f'(4) = -1\), so it is equivalent. 2. Statement 2 says \(f(4) = -1\), which is a function value rather than a derivative value, so it is not equivalent. 3. The derivative at a point is the tangent-line slope, so Statement 3 is equivalent. 4. Statement 4 is the limit definition of \(f'(4)\), so it is equivalent. 5. The line \(y = -x+10\) has slope \(-1\). Therefore, Statement 5 is equivalent. 6. Statement 6 gives a different function value and is not equivalent.

Statements 1, 3, 4, and 5 are equivalent to \(f'(4) = -1\).

- Average velocity is the slope of a secant line. - Instantaneous velocity is the limit of average velocities over shrinking intervals. - Simplify the difference quotient before taking the limit. - Distinguish between change over an interval and change at one instant.

1. Evaluate the positions: \(s(1) = 2(1)^2+5(1) = 7\,\text{m}\) and \(s(3) = 2(3)^2+5(3) = 33\,\text{m}\). 2. The average velocity on \([1, 3]\) is \(\frac{s(3)-s(1)}{3-1} = \frac{33-7}{2} = 13\,\frac{\text{m}}{\text{s}}\). 3. For the instantaneous velocity at \(t = 1\), simplify the difference quotient: \(\frac{s(1+h)-s(1)}{h} = \frac{2(1+h)^2+5(1+h)-7}{h} = \frac{9h+2h^2}{h} = 9+2h\). 4. Take the limit: \(s'(1) = \lim_{h\to 0}(9+2h) = 9\,\frac{\text{m}}{\text{s}}\).

Average velocity on \([1, 3]\): \(13\,\frac{\text{m}}{\text{s}}\) Instantaneous velocity at \(t = 1\): \(9\,\frac{\text{m}}{\text{s}}\)

- Substitute \(2+h\) for every \(x\) in the function. - Expand the squared binomial carefully. - Pay close attention to the negative sign before the square. - Factor and cancel \(h\) before taking the limit.

1. Evaluate \(f(2)\): \(f(2) = -(2)^2+6(2) = 8\). 2. Form the difference quotient: \(\frac{f(2+h)-f(2)}{h} = \frac{-(2+h)^2+6(2+h)-8}{h}\). 3. Simplify the numerator: \(-(4+4h+h^2)+12+6h-8 = 2h-h^2\). 4. For \(h \ne 0\), the quotient simplifies to \(\frac{2h-h^2}{h} = 2-h\). 5. Therefore, \(f'(2) = \lim_{h\to 0}(2-h) = 2\).

a) \(2-h\) b) \(f'(2) = 2\)

- Substitute \(3+h\) into the position function and expand. - What happens to the simplified expression as \(h\) approaches \(0\)? - Think about how a secant line approaches a tangent line.

1. The difference quotient is \(\frac{s(3+h)-s(3)}{h} = \frac{0.8(3+h)^2-0.8(3)^2}{h}\). Expanding and simplifying gives \(\frac{4.8h+0.8h^2}{h} = 4.8+0.8h\). 2. The instantaneous velocity is \(s'(3) = \lim_{h\to 0}(4.8+0.8h) = 4.8\,\frac{\text{m}}{\text{s}}\). 3. The instantaneous rate of change at \(t_0\) is the limit of the average rate of change on \([t_0, t_0+h]\) as the interval width \(h\) approaches \(0\). Geometrically, the secant line approaches the tangent line.

1. \(4.8+0.8h\) 2. \(4.8\,\frac{\text{m}}{\text{s}}\) 3. The instantaneous rate is the limit of the average rate over intervals whose width approaches \(0\).

- Start with the limit definition \(f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\). - Expand \((x+h)^2\). - Simplify the numerator until you can divide every term by \(h\). - Use the meaning of \(f'(x_0)\) as the slope at \(x_0\).

1. Set up the difference quotient: \(\frac{f(x+h)-f(x)}{h} = \frac{1.5(x+h)^2-1.5x^2}{h}\). 2. Expand and simplify: \(\frac{1.5(x^2+2xh+h^2)-1.5x^2}{h} = \frac{3xh+1.5h^2}{h}\). 3. For \(h\ne 0\), divide by \(h\): \(3x+1.5h\). 4. Take the limit as \(h\to 0\): \(f'(x) = \lim_{h\to 0}(3x+1.5h) = 3x\). 5. Set the derivative equal to the required slope: \(3x_0=9\). 6. Solve to get \(x_0=3\).

The derivative is \(f'(x) = 3x\). The graph has slope \(9\) at \(x_0=3\).

- What must be true about the slopes of parallel lines? - Be careful with signs when subtracting \(f(x)\) in the numerator. - Simplify the numerator so that you can divide by \(h\). - After dividing by \(h\), determine what happens to the remaining \(h\)-term as \(h\to 0\).

1. Set up the difference quotient: \(\frac{f(x+h)-f(x)}{h} = \frac{4(x+h)-(x+h)^2-(4x-x^2)}{h}\). 2. Expand the numerator: \(\frac{4x+4h-(x^2+2xh+h^2)-4x+x^2}{h}\). 3. Combine like terms: \(\frac{4h-2xh-h^2}{h}\). 4. For \(h\ne 0\), divide by \(h\), and then take the limit: \(f'(x)=\lim_{h\to 0}(4-2x-h)=4-2x\). 5. Parallel lines have equal slopes. Since \(y=-2x+5\) has slope \(-2\), solve \(4-2x_0=-2\). 6. This gives \(-2x_0=-6\), so \(x_0=3\).

The derivative is \(f'(x)=4-2x\). The tangent line is parallel to the given line at \(x_0=3\).

- Write the difference quotient for a general input \(x\). - Substitute \(x+h\) into the function and expand \((x+h)^2\). - Identify which terms cancel after subtracting \(f(x)\). - Simplify the quotient before taking the limit as \(h\to 0\).

1. Set up the difference quotient: \(\frac{f(x+h)-f(x)}{h} = \frac{5-2(x+h)^2-(5-2x^2)}{h}\). 2. Expand and simplify the numerator: \(\frac{5-2(x^2+2xh+h^2)-5+2x^2}{h} = \frac{-4xh-2h^2}{h}\). 3. For \(h\ne 0\), divide by \(h\): \(-4x-2h\). 4. Take the limit as \(h\to 0\): \(f'(x)=\lim_{h\to 0}(-4x-2h)=-4x\).

\(f'(x)=-4x\)

- Replace each occurrence of \(x\) in the function rule with \(x+h\). - Expand all expressions in the numerator and combine like terms. - Factor out \(h\) from the numerator so that you can simplify the quotient. - Determine what happens to the remaining \(h\)-term as \(h\to 0\).

1. Set up the difference quotient: \(\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2+6(x+h)-(x^2+6x)}{h}\). 2. Expand and combine like terms: \(\frac{x^2+2xh+h^2+6x+6h-x^2-6x}{h}=\frac{2xh+h^2+6h}{h}\). 3. For \(h\ne 0\), divide by \(h\): \(2x+h+6\). 4. Take the limit as \(h\to 0\): \(f'(x)=\lim_{h\to 0}(2x+h+6)=2x+6\).

\(f'(x)=2x+6\)

- Expand the squared binomial in the first method. - Factor the difference of squares in the second method. - A denominator of zero makes the original quotient undefined. - Compare how each variable point approaches \(x = 1\).

1. Using an increment \(h\), \(\frac{f(1+h)-f(1)}{h} = \frac{2(1+h)^2-2}{h} = \frac{4h+2h^2}{h} = 4+2h\) for \(h\ne 0\). Therefore, \(f'(1) = \lim_{h\to 0}(4+2h) = 4\). 2. Using a variable point \(x\), \(\frac{f(x)-f(1)}{x-1} = \frac{2x^2-2}{x-1} = \frac{2(x-1)(x+1)}{x-1} = 2(x+1)\) for \(x\ne 1\). Therefore, \(f'(1) = \lim_{x\to 1}2(x+1) = 4\). 3. Both methods produce the same tangent slope. Direct substitution into the original quotients gives \(\frac{0}{0}\), an undefined expression. Simplification for nearby values and then taking a limit avoids dividing by zero.

1. \(f'(1) = 4\) 2. \(f'(1) = 4\) 3. Direct substitution gives the undefined form \(\frac{0}{0}\); the limit must be taken after simplifying the quotient for nonzero separation.

- Expand \((4+h)^3\) carefully. - Simplify the difference quotient before taking the limit. - Terms that still contain \(h\) approach zero as \(h\to 0\). - The rate unit is volume divided by time.

1. Evaluate \(V(4) = 0.5(4)^3+2(4) = 40\). 2. Expand \(V(4+h)\): \(V(4+h) = 0.5(4+h)^3+2(4+h) = 40+26h+6h^2+0.5h^3\). 3. The difference quotient is \(\frac{V(4+h)-V(4)}{h} = \frac{26h+6h^2+0.5h^3}{h} = 26+6h+0.5h^2\), for \(h\ne 0\). 4. Taking the limit gives \(V'(4) = \lim_{h\to 0}(26+6h+0.5h^2) = 26\,\frac{\text{L}}{\text{min}}\). 5. At exactly \(4\) minutes, the tank’s water volume is increasing at \(26\,\frac{\text{L}}{\text{min}}\).

a) \(26+6h+0.5h^2\) b) \(26\,\frac{\text{L}}{\text{min}}\) c) At \(t = 4\), the water volume is increasing at \(26\,\frac{\text{L}}{\text{min}}\).

- Start with the definition of instantaneous rate of change as a limit. - Expand \((a+b)^3\). - Simplify the quotient so that \(h\) no longer appears in the denominator. - Determine what happens to terms containing \(h\) as \(h\to 0\).

1. Set up the difference quotient: \(\frac{f(x_0+h)-f(x_0)}{h}=\frac{(x_0+h)^3-x_0^3}{h}\). 2. Expand the cube: \(\frac{x_0^3+3x_0^2h+3x_0h^2+h^3-x_0^3}{h}\). 3. Simplify the numerator: \(\frac{3x_0^2h+3x_0h^2+h^3}{h}\). 4. For \(h\ne 0\), divide by \(h\): \(3x_0^2+3x_0h+h^2\). 5. Take the limit as \(h\to 0\): \(f'(x_0)=\lim_{h\to 0}(3x_0^2+3x_0h+h^2)=3x_0^2\). Therefore, \(f'(x)=3x^2\).

\(f'(x_0)=3x_0^2\), so \(f'(x)=3x^2\).

- Substitute the function and point into the difference quotient. - Factor the numerator so that the denominator can be canceled. - Use a difference of squares more than once. - Evaluate the simplified expression at \(x = 2\).

1. Form the difference quotient: \(\frac{f(x)-f(2)}{x-2} = \frac{x^4-16}{x-2}\). 2. Factor the numerator: \(x^4-16 = (x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4)\). 3. For \(x\ne 2\), cancel \(x-2\) to obtain \((x+2)(x^2+4)\). 4. Take the limit: \(f'(2) = \lim_{x\to 2}(x+2)(x^2+4) = (4)(8) = 32\).

The tangent-line slope is \(f'(2) = 32\).

- Substitute the point into the limit definition. - Use a common denominator to subtract the fractions in the numerator. - Be careful with the negative sign when simplifying. - Cancel \(h\) before taking the limit.

1. Evaluate \(f(1) = \frac{1}{2}\) and \(f(1+h) = \frac{1}{2+h}\). 2. Form the difference quotient: \(\frac{f(1+h)-f(1)}{h} = \frac{\frac{1}{2+h}-\frac{1}{2}}{h}\). 3. Combine the fractions in the numerator: \(\frac{1}{2+h}-\frac{1}{2} = \frac{2-(2+h)}{2(2+h)} = \frac{-h}{2(2+h)}\). 4. For \(h\ne 0\), the quotient simplifies to \(-\frac{1}{2(2+h)}\). 5. Take the limit: \(f'(1) = \lim_{h\to 0}\left(-\frac{1}{2(2+h)}\right) = -\frac{1}{4}\).

\(f'(1) = -\frac{1}{4} = -0.25\)

- Combine the two fractions in the numerator using a common denominator. - Simplify the resulting complex fraction before taking the limit. - The factor \(h\) must cancel from the difference quotient. - Substitute the limiting value only after simplifying.

1. Set up the difference quotient: \(\frac{f(a+h)-f(a)}{h}=\frac{\frac{3}{a+h}-\frac{3}{a}}{h}\). 2. Combine the fractions in the numerator: \(\frac{3}{a+h}-\frac{3}{a}=\frac{3a-3(a+h)}{a(a+h)}=\frac{-3h}{a(a+h)}\). 3. For \(h\ne 0\), simplify the difference quotient: \(\frac{-3h}{a(a+h)}\cdot\frac{1}{h}=\frac{-3}{a(a+h)}\). 4. Take the limit as \(h\to 0\): \(f'(a)=\lim_{h\to 0}\frac{-3}{a(a+h)}=-\frac{3}{a^2}\).

\(f'(a)=-\frac{3}{a^2}\)

- Write the difference quotient using \(f(1+h)\) and \(f(1)\). - Expand \((1+h)^2\) carefully. - Factor out and cancel \(h\) before taking the limit. - Evaluate the simplified expression as \(h\) approaches \(0\).

1. Evaluate the function at the point: \(f(1) = 2(1)^2-1 = 1\). 2. Form the difference quotient: \(\frac{f(1+h)-f(1)}{h} = \frac{2(1+h)^2-(1+h)-1}{h}\). 3. Expand and simplify the numerator: \(2(1+2h+h^2)-1-h-1 = 3h+2h^2\). 4. For \(h\ne 0\), simplify the quotient: \(\frac{3h+2h^2}{h} = 3+2h\). 5. Take the limit: \(f'(1) = \lim_{h\to 0}(3+2h) = 3\).

The slope of the tangent line at \(x = 1\) is \(3\).

- Expand \((2+h)^3\) before simplifying the difference quotient. - Distribute the factor \(\frac{1}{2}\) to every term. - After subtracting \(f(2)\), each term in the numerator should contain \(h\). - Cancel \(h\), then take the limit as \(h\) approaches \(0\).

1. Evaluate the function at the point: \(f(2) = \frac{1}{2}(2)^3+1 = 5\). 2. Form the difference quotient: \(\frac{f(2+h)-f(2)}{h} = \frac{\frac{1}{2}(2+h)^3+1-5}{h}\). 3. Expand the cube: \((2+h)^3 = 8+12h+6h^2+h^3\). 4. Simplify the numerator: \(\frac{1}{2}(8+12h+6h^2+h^3)-4 = 6h+3h^2+\frac{1}{2}h^3\). 5. For \(h\ne 0\), simplify the quotient: \(6+3h+\frac{1}{2}h^2\). 6. Take the limit: \(f'(2) = \lim_{h\to 0}\left(6+3h+\frac{1}{2}h^2\right) = 6\).

The instantaneous rate of change at \(x = 2\) is \(6\).

- Begin with average velocity over a short time interval. - Let the length of that interval approach \(0\). - Substitute the position function into the difference quotient. - Simplify so that the factor \(h\) can be canceled before taking the limit.

1. The average velocity from \(t = 4\) to \(t = 4+h\) is \(\frac{s(4+h)-s(4)}{h}\). 2. Evaluate the position values: \(s(4) = 0.5(4)^2 = 8\) and \(s(4+h) = 0.5(4+h)^2 = 8+4h+0.5h^2\). 3. Simplify the difference quotient: \(\frac{8+4h+0.5h^2-8}{h} = 4+0.5h\) for \(h\ne 0\). 4. Define and evaluate the instantaneous velocity: \(v(4) = \lim_{h\to 0}\frac{s(4+h)-s(4)}{h} = \lim_{h\to 0}(4+0.5h) = 4\). 5. Therefore, the instantaneous velocity at \(t = 4\) is \(4\,\frac{\text{m}}{\text{s}}\).

\(v(4) = \lim_{h\to 0}\frac{s(4+h)-s(4)}{h} = 4\,\frac{\text{m}}{\text{s}}\)

- Interpret change in volume divided by change in time. - Track the units of the quotient. - Expand \((5+h)^2\) and simplify the numerator. - Cancel \(h\) before letting \(h\) approach \(0\).

1. The difference quotient is \(\frac{V(5+h)-V(5)}{h}\). It represents the average rate of change of the water volume from \(5\) to \(5+h\) minutes, measured in liters per minute. 2. Evaluate \(V(5) = 10(5)-0.2(5)^2 = 45\). 3. Evaluate and simplify \(V(5+h) = 10(5+h)-0.2(5+h)^2 = 45+8h-0.2h^2\). 4. Simplify the difference quotient: \(\frac{45+8h-0.2h^2-45}{h} = 8-0.2h\) for \(h\ne 0\). 5. Take the limit: \(V'(5) = \lim_{h\to 0}(8-0.2h) = 8\). 6. The volume is increasing at an instantaneous rate of \(8\,\frac{\text{L}}{\text{min}}\) when \(t = 5\).

The quotient \(\frac{V(5+h)-V(5)}{h}\) is the average change in volume per minute from \(t = 5\) to \(t = 5+h\). The instantaneous rate at \(t = 5\) is \(8\,\frac{\text{L}}{\text{min}}\).

- Use the slope formula with the two endpoint values for the average rate. - For the instantaneous rate, let a second time approach \(t = 1\). - A secant line passes through two points on a graph; a tangent line gives the local slope at one point. - Rate units are volume per unit of time.

1. Evaluate the endpoints: \(V(1) = 1^2+5(1) = 6\) and \(V(4) = 4^2+5(4) = 36\). 2. The average rate of change is \(\frac{V(4)-V(1)}{4-1} = \frac{36-6}{3} = 10\,\frac{\text{L}}{\text{min}}\). 3. For the instantaneous rate, form the difference quotient: \(\frac{V(1+h)-V(1)}{h} = \frac{(1+h)^2+5(1+h)-6}{h}\). 4. Expand and simplify: \(\frac{7h+h^2}{h} = 7+h\) for \(h\ne 0\). 5. Take the limit: \(V'(1) = \lim_{h\to 0}(7+h) = 7\,\frac{\text{L}}{\text{min}}\). 6. The average rate is the slope of the secant line through \((1, 6)\) and \((4, 36)\). The instantaneous rate is the slope of the tangent line at \((1, 6)\).

1. \(10\,\frac{\text{L}}{\text{min}}\) 2. \(7\,\frac{\text{L}}{\text{min}}\) 3. The average rate is a secant-line slope, and the instantaneous rate is the tangent-line slope at \(t = 1\).

- Average velocity is displacement divided by elapsed time. - Instantaneous velocity is the derivative of position. - Substitute \(t = 2+h\) and \(t = 2\) into the position function. - Think about what happens to average velocity as the time interval becomes arbitrarily short.

1. Evaluate the positions: \(s(0) = 0\) and \(s(5) = 30(5)-5^2 = 125\). 2. The average velocity is \(\frac{s(5)-s(0)}{5-0} = \frac{125}{5} = 25\,\frac{\text{m}}{\text{s}}\). 3. Form the difference quotient at \(t = 2\): \(\frac{s(2+h)-s(2)}{h}\). 4. Since \(s(2) = 56\), the quotient becomes \(\frac{30(2+h)-(2+h)^2-56}{h} = \frac{26h-h^2}{h} = 26-h\) for \(h\ne 0\). 5. Take the limit: \(s'(2) = \lim_{h\to 0}(26-h) = 26\,\frac{\text{m}}{\text{s}}\). 6. In general, the instantaneous rate at \(t_0\) is the limit of the average rate on \([t_0,t_0+h]\) as \(h\) approaches \(0\).

1. \(25\,\frac{\text{m}}{\text{s}}\) 2. \(26\,\frac{\text{m}}{\text{s}}\) 3. The instantaneous rate is the limit of the average rate as the interval length approaches \(0\).

- Substitute \(100+h\) and \(100\) carefully into the cost function. - The constant terms in the numerator should cancel. - Cancel \(h\) before taking the limit. - Marginal cost approximates the cost of producing one additional unit.

1. Form the difference quotient: \(\frac{C(100+h)-C(100)}{h}\). 2. Substitute the cost function: \(\frac{0.05(100+h)^2+10(100+h)+500-(0.05(100)^2+10(100)+500)}{h}\). 3. Expanding and combining like terms gives \(\frac{20h+0.05h^2}{h} = 20+0.05h\) for \(h\ne 0\). The quotient has units of dollars per unit. 4. Take the limit: \(C'(100) = \lim_{h\to 0}(20+0.05h) = 20\,\frac{\$}{\text{unit}}\). 5. Therefore, the model estimates that producing the 101st unit adds approximately \(\$20\) to total cost.

1. \((20+0.05h)\,\frac{\$}{\text{unit}}\) 2. \(C'(100) = 20\,\frac{\$}{\text{unit}}\) 3. The 101st unit adds approximately \(\$20\) to total cost.

- Interpret the derivative as the limiting slope of secant lines whose points move together. - Combine the two fractions in the numerator using a common denominator. - Factor the numerator so that \(x-a\) can be canceled. - Evaluate the remaining expression as \(x\to a\).

1. Set up the difference quotient as \(x\to a\): \(\frac{f(x)-f(a)}{x-a}=\frac{\frac{1}{x^2}-\frac{1}{a^2}}{x-a}\). 2. Combine the fractions in the numerator: \(\frac{a^2-x^2}{x^2a^2(x-a)}\). 3. Factor the numerator: \(\frac{-(x^2-a^2)}{x^2a^2(x-a)}=\frac{-(x-a)(x+a)}{x^2a^2(x-a)}\). 4. For \(x\ne a\), cancel \(x-a\): \(-\frac{x+a}{x^2a^2}\). 5. Take the limit: \(f'(a)=\lim_{x\to a}-\frac{x+a}{x^2a^2}=-\frac{2a}{a^4}=-\frac{2}{a^3}\).

\(f'(a)=-\frac{2}{a^3}\)

- Select two easy-to-read points on the tangent line. - Use rise divided by run for the graphical slope. - Substitute \(3+h\) into the function and simplify. - Cancel \(h\) before taking the limit.

1. The tangent line passes through \(A(3, 0)\) and the grid point \((4, 2)\). Its slope is \(\frac{2-0}{4-3} = 2\). 2. For the limit calculation, first find \(f(3) = 0\). 3. Evaluate \(f(3+h) = (3+h)^2-4(3+h)+3 = h^2+2h\). 4. Form and simplify the difference quotient: \(\frac{f(3+h)-f(3)}{h} = \frac{h^2+2h}{h} = h+2\) for \(h\ne 0\). 5. Take the limit: \(f'(3) = \lim_{h\to 0}(h+2) = 2\). Both methods give the same instantaneous rate of change.

a) \(2\) b) \(f'(3) = 2\)

- Use the slope formula for each secant line. - Find \(f(3)\) before calculating the second slope. - Think about what the secant line becomes as its second point approaches \(P\). - Simplify the difference quotient before taking the limit.

1. The slope through \(P(2, 2)\) and \(Q(4, 5)\) is \(\frac{5-2}{4-2} = \frac{3}{2} = 1.5\). 2. Evaluate \(f(3) = 0.25(3)^2+1 = 3.25\), so \(Q_1 = (3, 3.25)\). 3. The second secant slope is \(\frac{3.25-2}{3-2} = 1.25\). 4. The secant slopes decrease toward \(1\) as the second point approaches \(P\), so predict \(f'(2) = 1\). 5. Evaluate \(f(2+h) = 0.25(2+h)^2+1 = 2+h+0.25h^2\). 6. The difference quotient is \(\frac{f(2+h)-f(2)}{h} = \frac{h+0.25h^2}{h} = 1+0.25h\) for \(h\ne 0\). 7. Take the limit: \(f'(2) = \lim_{h\to 0}(1+0.25h) = 1\).

a) \(1.5\) b) \(1.25\) c) Prediction: \(1\); verification: \(f'(2) = 1\)

- Find \(f(1)\) first. - Substitute \(1+h\) into the function and simplify. - Factor out and cancel \(h\) in the difference quotient. - Evaluate the remaining expression as \(h\) approaches \(0\).

1. Evaluate \(f(1) = 1^2+4(1) = 5\). 2. Evaluate \(f(1+h) = (1+h)^2+4(1+h) = h^2+6h+5\). 3. Form and simplify the difference quotient: \(\frac{f(1+h)-f(1)}{h} = \frac{h^2+6h}{h} = h+6\) for \(h\ne 0\). 4. Take the limit: \(f'(1) = \lim_{h\to 0}(h+6) = 6\).

\(f'(1) = 6\)

- Use \(\frac{f(x_0+h)-f(x_0)}{h}\) with \(x_0 = -1\). - Expand \((-1+h)^2\) carefully. - Cancel \(h\) before taking the limit.

1. Evaluate \(f(-1) = 3(-1)^2-2 = 1\). 2. Evaluate \(f(-1+h) = 3(-1+h)^2-2 = 3h^2-6h+1\). 3. Form and simplify the difference quotient: \(\frac{f(-1+h)-f(-1)}{h} = \frac{3h^2-6h}{h} = 3h-6\) for \(h\ne 0\). 4. Take the limit: \(f'(-1) = \lim_{h\to 0}(3h-6) = -6\).

The tangent-line slope is \(-6\).

- Compare the left-hand and right-hand limits at \(0\). - Write the slope between \(P(0, 0)\) and a point \((x,s(x))\) on the graph. - Determine whether those secant slopes approach a finite value as \(x\to0\).

1. The graph has a jump at \(x=0\): it lies at \(y=-1\) to the left and \(y=1\) to the right, while \(s(0)=0\). 2. The one-sided limits are \(\lim_{x\to0^-}s(x)=-1\) and \(\lim_{x\to0^+}s(x)=1\). Since they differ, \(s\) is not continuous at \(0\). 3. For \(x\ne0\), the secant slope through \(P(0, 0)\) and \((x,s(x))\) is \(\frac{s(x)-s(0)}{x-0}=\frac{s(x)}{x}=\frac{1}{|x|}\). As \(x\to0\), these slopes increase without bound, so they do not approach a finite number. Therefore, \(s^{\prime}(0)\) does not exist.

1) The graph has a jump at \(x=0\), with \(s(0)=0\). 2) \(\lim_{x\to0^-}s(x)=-1\) and \(\lim_{x\to0^+}s(x)=1\), so \(s\) is not continuous at \(0\). 3) The secant slopes equal \(\frac{1}{|x|}\), which increases without bound as \(x\to0\). Thus, \(s^{\prime}(0)\) does not exist as a finite number.

- Substitute \(x+h\) for every occurrence of \(x\) in the function rule. - Expand \((x+h)^2\). - Simplify the numerator until every remaining term contains a factor of \(h\). - Cancel \(h\) before taking the limit.

1. Set up the difference quotient: \(\frac{f(x+h)-f(x)}{h}=\frac{0.5(x+h)^2+2-(0.5x^2+2)}{h}\). 2. Expand and simplify the numerator: \(\frac{0.5(x^2+2xh+h^2)+2-0.5x^2-2}{h}=\frac{xh+0.5h^2}{h}\). 3. For \(h\ne 0\), factor and divide by \(h\): \(\frac{h(x+0.5h)}{h}=x+0.5h\). 4. Take the limit as \(h\to 0\): \(f'(x)=\lim_{h\to 0}(x+0.5h)=x\).

\(f'(x)=x\)

- Expand \((t+h)^2\) and combine like terms in the difference quotient. - All terms without \(h\) should cancel from the numerator. - The sign of the derivative tells whether the temperature is increasing or decreasing. - To answer part b, set the derivative equal to the given rate.

1. Form and simplify the difference quotient: \(\frac{T(t+h)-T(t)}{h} = \frac{0.2(t+h)^2-8(t+h)+100-(0.2t^2-8t+100)}{h}\). 2. Expanding and combining like terms gives \(\frac{0.4th+0.2h^2-8h}{h} = 0.4t+0.2h-8\) for \(h\ne 0\). 3. Take the limit: \(T'(t) = \lim_{h\to 0}(0.4t+0.2h-8) = 0.4t-8\). 4. Evaluate the rates: \(T'(5) = 0.4(5)-8 = -6\) and \(T'(15) = 0.4(15)-8 = -2\). The units are degrees Celsius per minute. 5. Solve \(0.4t-8 = -2\): \(0.4t = 6\), so \(t = 15\) minutes. 6. For \(0\le t<20\), \(0.4t-8<0\), so the temperature is decreasing. At \(t = 20\), the rate is zero, so the temperature has stopped decreasing instantaneously at the endpoint of the model.

a) \(T'(5) = -6\,\frac{{}^\circ\text{C}}{\text{min}}\) and \(T'(15) = -2\,\frac{{}^\circ\text{C}}{\text{min}}\) b) \(t = 15\,\text{min}\) c) The negative rate means the temperature is decreasing. At \(t = 20\), the instantaneous rate is zero, so the temperature has stopped decreasing instantaneously at the endpoint of the model.

- Write the difference quotient for the function. - Simplify the complex fraction. - Use the given identity to factor out the term \(x-a\). - After canceling, evaluate the limit as \(x\to a\).

1. Set up the difference quotient: \(\frac{f(x)-f(a)}{x-a}=\frac{\frac{1}{x^4}-\frac{1}{a^4}}{x-a}\). 2. Combine the fractions in the numerator: \(\frac{a^4-x^4}{x^4a^4(x-a)}\). 3. Factor the numerator: \(\frac{-(x^4-a^4)}{x^4a^4(x-a)}=\frac{-(x-a)(x^3+x^2a+xa^2+a^3)}{x^4a^4(x-a)}\). 4. For \(x\ne a\), cancel \(x-a\): \(-\frac{x^3+x^2a+xa^2+a^3}{x^4a^4}\). 5. As \(x\to a\), the numerator approaches \(-4a^3\) and the denominator approaches \(a^8\). 6. Therefore, \(f'(a)=-\frac{4a^3}{a^8}=-\frac{4}{a^5}\).

\(f'(a)=-\frac{4}{a^5}\)