Free math worksheets

- Recall the continuity property of polynomial functions. - Evaluate the function at the target input. - Compare the limiting value with the actual function value.

1. A polynomial is continuous at every real input, so its limit can be found by direct substitution. 2. \(\lim_{x\to-1}P(x)=(-1)^4-3\cdot(-1)+2=6\). 3. Also \(P(-1)=6\). Since the limit equals the function value, \(P\) is continuous at \(x=-1\).

\(P\) is continuous at \(x=-1\), and both the limit and function value equal \(6\).

- Find the domain of the logarithm before discussing continuity. - Use continuity of a composition on the allowed inputs. - Check whether the boundary input itself belongs to the domain.

1. The logarithm requires \(x-1>0\), so the domain is \((1, \infty)\). 2. The function \(x-1\) is continuous, and \(\ln u\) is continuous for \(u>0\), so their composition is continuous throughout that domain. 3. The input \(x=1\) is excluded because \(\ln(0)\) is undefined.

\(L\) is continuous on \((1, \infty)\).

- Evaluate the two formulas through one-sided limits at the joining point. - Identify which branch supplies the actual value at \(0\). - Continuity requires all three quantities to agree.

1. The left-hand limit is \(e^0=1\). 2. The right-hand limit is \(0+1=1\), so the two-sided limit is \(1\). 3. Because the left branch includes \(0\), \(E(0)=e^0=1\). The limit equals the function value, so \(E\) is continuous at \(0\).

\(E\) is continuous at \(x=0\); both one-sided limits and \(E(0)\) equal \(1\).

- Analyze the inner function before the logarithm. - Check whether the inner output can ever leave the logarithm's domain. - Use the composition continuity rule once the domain condition is settled.

1. The inner function \(x^2+1\) is a polynomial, so it is continuous for all real \(x\). 2. Also \(x^2+1\ge1>0\) for every real \(x\), so its values always lie in the logarithm's domain. 3. Therefore the composition \(\ln(x^2+1)\) is continuous on all real numbers.

\(J\) is continuous on \(\mathbb{R}\).

- Identify the continuity of the numerator and denominator separately. - Apply the quotient continuity rule with its denominator restriction. - Use the excluded input to split the real line into intervals.

1. The sine function and the linear function \(x+2\) are continuous for all real inputs. 2. Their quotient is continuous wherever the denominator is nonzero. 3. Since \(x+2=0\) only at \(x=-2\), \(B\) is continuous on \((-\infty, -2)\) and \((-2, \infty)\).

\((-\infty, -2)\) and \((-2, \infty)\).

- Evaluate the function’s behavior from each side of the joining point. - Use the correct piece for the actual value at \(x=0\). - Continuity requires both one-sided limits and the function value to agree.

1. From the left, \(\lim_{x\to0^-}g(x)=\lim_{x\to0^-}(x^2+1)=1\). 2. Since the second rule applies at \(x=0\), \(g(0)=2-0=2\). The right-hand limit is also \(2\). 3. The one-sided limits are different, and the left-hand limit does not equal the function value. Therefore, the two-sided limit does not exist and \(g\) is discontinuous at \(0\).

\(g\) is discontinuous at \(x=0\) because \(\lim_{x\to0^-}g(x)=1\), while \(\lim_{x\to0^+}g(x)=g(0)=2\).

- Check each joining point separately. - Compare the left-hand limit, right-hand limit, and function value. - Use the interval conditions to select the correct rule at each endpoint.

1. At \(x=2\), the left-hand limit is \(2^2-1=3\). The middle piece gives the right-hand limit and function value \(f(2)=1.5\cdot 2=3\). Therefore, \(f\) is continuous at \(2\). 2. At \(x=4\), the left-hand limit from the middle piece is \(1.5\cdot 4=6\). The final piece gives the right-hand limit and function value \(f(4)=\sqrt{4}+4=6\). Therefore, \(f\) is continuous at \(4\).

The function is continuous at both joining points. The common value is \(3\) at \(x=2\) and \(6\) at \(x=4\).

- Imagine tracing the graph from left to right without lifting your pencil. - Distinguish a corner from a hole or jump. - At each input, compare what the graph approaches from both sides with the actual function value.

1. At \(x=-1\), the left-hand limit is \(2\), while the right-hand limit is \(0\). Since the one-sided limits are different, \(f\) has a jump discontinuity at \(x=-1\). 2. At \(x=1\), the graph follows the same smooth piece from both sides and passes through \((1, 2)\). Thus \(\lim_{x\to1}f(x)=f(1)=2\), so \(f\) is continuous there. 3. At \(x=2\), both pieces meet at \((2, 1.5)\). Therefore, the left-hand limit, right-hand limit, and function value all equal \(1.5\), so \(f\) is continuous there. The corner affects differentiability, not continuity.

a) Not continuous; there is a jump at \(x=-1\). b) Continuous at \(x=1\). c) Continuous at \(x=2\), even though the graph has a corner there.

- A change from a curve to a line does not necessarily cause a discontinuity. - Look for an isolated filled point that does not lie on the value approached by the graph.

1. At \(x=0\), the curve approaches \(0\) from the left, and the horizontal piece approaches \(0\) from the right. The filled point gives \(p(0)=0\). Therefore, \(\lim_{x\to0}p(x)=p(0)=0\), so \(p\) is continuous at \(x=0\). 2. At \(x=2\), the graph approaches \(0\) from both sides, so \(\lim_{x\to2}p(x)=0\). However, the filled point gives \(p(2)=1\). Since the limit and function value are different, \(p\) has a removable discontinuity at \(x=2\).

At \(x=0\), \(p\) is continuous. At \(x=2\), \(p\) is not continuous; the discontinuity is removable.

- Check whether the graph approaches the same point from both sides. - Look for holes or jumps. - A corner does not by itself make a function discontinuous. - Compare the limiting value with the actual function value.

1. At \(x=0\), the two pieces meet at \((0, 2)\). The one-sided limits and \(h(0)\) are all \(2\), so \(h\) is continuous there. The corner does not affect continuity. 2. At \(x=2\), the left-hand limit and function value are \(1\), while the right-hand limit is \(3\). Since the one-sided limits are different, \(h\) has a jump discontinuity at \(x=2\). 3. At \(x=4\), the graph lies on one unbroken line segment, so the limit equals \(h(4)=4\). Therefore, \(h\) is continuous at \(x=4\).

At \(x=0\), \(h\) is continuous. At \(x=2\), \(h\) is not continuous; it has a jump discontinuity. At \(x=4\), \(h\) is continuous.

- Start with the general continuity rule for rational functions. - Find every input that makes the denominator zero. - Use those excluded inputs to split the real line into continuity intervals.

1. A rational function is continuous wherever its denominator is nonzero. 2. The denominator factors as \(x^2-9=(x-3)(x+3)\), so it is zero at \(x=-3\) and \(x=3\). 3. Therefore \(R\) is continuous on \((-\infty, -3)\), \((-3, 3)\), and \((3, \infty)\).

\((-\infty, -3)\), \((-3, 3)\), and \((3, \infty)\).

- Find the real domain before discussing continuity. - Interior points and endpoints use slightly different limit conditions. - At the left endpoint of a domain interval, only nearby domain inputs to the right are available.

1. The square root requires \(x+2\ge0\), so the real domain is \([-2, \infty)\). 2. The function is continuous at every interior point of this domain. 3. At the endpoint, \(\lim_{x\to-2^+}\sqrt{x+2}=0=S(-2)\), so \(S\) is continuous on \([-2, \infty)\) using endpoint continuity.

\(S\) is continuous on \([-2, \infty)\). At \(x=-2\), continuity is checked with the right-hand limit because the domain has nearby points only to the right.

- Solve the inequality that makes the radicand nonnegative. - Use continuity of a composition on interior points of the domain. - At each endpoint, approach only through points that remain in the domain.

1. The radicand must satisfy \(4-x^2\ge0\), so the domain is \([-2, 2]\). 2. On the interior \((-2, 2)\), the polynomial \(4-x^2\) is continuous and positive, and square root is continuous on nonnegative inputs, so the composition is continuous. 3. At \(-2\), the right-hand limit equals \(H(-2)=0\); at \(2\), the left-hand limit equals \(H(2)=0\). Thus \(H\) is continuous on \([-2, 2]\).

\(H\) is continuous on \([-2, 2]\), using right-hand continuity at \(-2\) and left-hand continuity at \(2\).

- Translate continuity of \(f\) and \(g\) into limit statements. - Recall the condition needed for the quotient law. - Finish by comparing the quotient limit with the quotient's point value.

1. Continuity gives \(\lim_{x\to a}f(x)=f(a)\) and \(\lim_{x\to a}g(x)=g(a)\). 2. Since \(g(a)\ne0\), the quotient law applies, so \(\lim_{x\to a}Q(x)=\frac{f(a)}{g(a)}\). 3. This equals \(Q(a)\), so the limit equals the function value and \(Q\) is continuous at \(a\).

\(Q\) is continuous at \(x=a\) because the quotient limit exists and equals \(Q(a)=\frac{f(a)}{g(a)}\), with a nonzero denominator.

- Read the y-value approached from the left and from the right. - Compare the common limiting value with the filled point at \(x=3\). - Continuity does not require the graph to be smooth.

1. From both sides of \(x=3\), the graph approaches the y-value \(0\). 2. The filled point is \((3,0)\), so \(A(3)=0\). 3. Since \(\lim_{x\to3}A(x)=A(3)=0\), the function is continuous at \(x=3\). The corner affects differentiability, not continuity.

\(A\) is continuous at \(x=3\) because both the limit and the function value equal \(0\).

- Separate the special point \(x=0\) from all other inputs. - Use the quotient continuity rule where the denominator is nonzero. - At the special point, compare the known limit with the assigned value.

1. For every \(x\ne0\), \(S(x)=\frac{\sin x}{x}\) is a quotient of continuous functions with nonzero denominator, so it is continuous. 2. At \(x=0\), the standard limit \(\lim_{x\to0}\frac{\sin x}{x}=1\). 3. Since \(S(0)=1\), the limit equals the function value at the only point needing a separate check. Therefore \(S\) is continuous on \(\mathbb{R}\).

\(S\) is continuous for every real \(x\).

- Separate interior continuity from endpoint continuity. - At each endpoint, approach only through inputs that lie inside the interval. - Match each one-sided limit with the function value at that endpoint.

1. Continuity on the interior \((0, 3)\) is already given. 2. At the left endpoint, the right-hand limit equals \(f(0)=2\). 3. At the right endpoint, the left-hand limit equals \(f(3)=-1\). These are the endpoint continuity conditions, so \(f\) is continuous on \([0, 3]\). A left-hand limit at \(0\) and a right-hand limit at \(3\) are not required for continuity on this closed interval.

Yes. \(f\) is continuous on \([0, 3]\) because it is continuous on the interior and satisfies the appropriate one-sided continuity conditions at both endpoints.

- Separate the idea of continuity from the behavior of slope. - Compare the limit at \(0\) with the function value there. - Recall the real domain of the cube-root function.

1. The cube-root function is defined and continuous for every real input. 2. In particular, \(\lim_{x\to0}\sqrt[3]{x}=0=C(0)\). 3. A vertical tangent concerns the derivative, not whether nearby function values approach the function value. Thus \(C\) is continuous on \(\mathbb{R}\).

Yes. \(C\) is continuous at \(0\) and on all real numbers; the vertical tangent does not create a discontinuity.

- Combine the domain condition for a square root with the restriction from division. - Determine whether the boundary input is actually in the function's domain. - Endpoint continuity can only include an endpoint where the function itself is defined.

1. The square root requires \(x-2\ge0\), but the result is also in the denominator, so it must be strictly positive. 2. Therefore \(x>2\), giving domain \((2, \infty)\). 3. On this interval the square-root and reciprocal operations are continuous. At \(x=2\), the denominator is \(0\), so the function is undefined and the endpoint cannot be included.

\(D\) is continuous on \((2, \infty)\).

- Start from what continuity of the inner function tells you about its limit. - Check whether the outer square-root function is continuous at that limiting input. - Compare the composed limit with the composed function value.

1. Continuity of \(f\) gives \(\lim_{x\to a}f(x)=9\). Since this limiting value is positive, \(f(x)\) remains positive for inputs sufficiently close to \(a\). 2. The square-root function is continuous at \(9\), so \(\lim_{x\to a}\sqrt{f(x)}=\sqrt{9}=3\). 3. Also \(G(a)=\sqrt{f(a)}=3\). Therefore the limit equals the function value and \(G\) is continuous at \(a\).

\(G(a)=3\), and \(G\) is continuous at \(x=a\).

- Rewrite tangent in terms of sine and cosine if needed. - Find where the denominator of that quotient is zero in the stated interval. - Use those excluded inputs to divide the interval into maximal continuous pieces.

1. The tangent function is continuous wherever \(\cos x\ne0\). 2. On \([0, 2\pi]\), \(\cos x=0\) at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). 3. Therefore the maximal continuity subintervals are \([0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \frac{3\pi}{2})\), and \((\frac{3\pi}{2}, 2\pi]\).

\([0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \frac{3\pi}{2})\), and \((\frac{3\pi}{2}, 2\pi]\).

- Find the left-hand limit using the first piece. - Evaluate the second piece at the joining point. - Set the two values equal and solve for the parameter.

1. The left-hand limit is \(\lim_{x\to-1^-}\frac{6}{x-2}=\frac{6}{-3}=-2\). 2. The second rule gives the function value and right-hand limit: \(g(-1)=a\cdot 2^2-10=4a-10\). 3. Continuity requires \(4a-10=-2\). 4. Solving gives \(4a=8\), so \(a=2\).

\(a=2\)

- Identify where a rational function can fail to be continuous. - Determine when \(x^2+c\) has no real zeros. - Test a simple positive value of \(c\).

1. A rational function is continuous at every point in its domain, so the only possible discontinuities occur where the denominator is \(0\). 2. The denominator equation is \(x^2+c=0\). If \(c>0\), then \(x^2+c>0\) for every real \(x\), so there are no excluded inputs. 3. For example, \(c=1\) gives \(g_1(x)=\frac{1}{x^2+1}\), which is defined and continuous for every real \(x\). 4. Therefore, the statement is true.

The statement is true. Every \(c>0\) makes \(x^2+c\) nonzero for all real \(x\), so \(g_c\) is continuous on \(\mathbb{R}\).

- Substitute the joining-point value into each neighboring piece. - Notice what happens to \(k(x-5)^2\) at \(x=5\). - Set the two expressions at \(x=8\) equal.

1. At \(x=5\), the first piece gives \(g(5)=k\cdot 0^2+2=2\). Its left-hand limit is also \(2\). 2. The second piece is constantly \(2\), so the right-hand limit at \(5\) is \(2\). Thus, \(g\) is continuous at \(5\) for every \(k\). 3. At \(x=8\), the second piece gives \(g(8)=2\) and the left-hand limit is \(2\). 4. The right-hand limit is \(1.5\cdot 8+d=12+d\). Continuity requires \(12+d=2\), so \(d=-10\).

a) Both one-sided limits and \(g(5)\) equal \(2\) for every \(k\). b) \(d=-10\)

- Select the piece containing each input. - At the threshold, compare the left-hand limit, right-hand limit, and function value. - Interpret continuity as the absence of a sudden jump.

1. The second piece applies at \(115\): \(T(115)=2\cdot(115-95)=40\). 2. The third piece applies at \(125\): \(T(125)=2.20\cdot(125-115)+40=22+40=62\). 3. At \(x=115\), the left-hand limit is \(2\cdot(115-95)=40\). The right-hand limit is \(2.20\cdot(115-115)+40=40\), and \(T(115)=40\). 4. Since all three values agree, \(T\) is continuous at \(115\). A very small change across the threshold does not create a sudden jump in the fee.

a) \(T(115)=\$40\) and \(T(125)=\$62\) b) \(T\) is continuous at \(x=115\), because both one-sided limits and the function value equal \(40\). There is no abrupt fee increase at the threshold.

- Evaluate the neighboring pieces at each joining point. - Use the interval conditions to determine the actual function value. - A jump occurs when the one-sided limits are different.

1. At \(x=-2\), the left-hand limit is \(-2+5=3\). The right-hand limit is \((-2)^2-1=3\), and \(g(-2)=3\). Therefore, \(g\) is continuous at \(-2\). 2. At \(x=3\), the left-hand limit is \(3^2-1=8\). The right-hand limit is \(\frac{18}{3}=6\), while \(g(3)=8\). 3. Since the one-sided limits at \(3\) are different, the two-sided limit does not exist. The function has a jump discontinuity at \(3\).

At \(x=-2\): left-hand limit \(=3\), right-hand limit \(=3\), and \(g(-2)=3\); continuous. At \(x=3\): left-hand limit \(=8\), right-hand limit \(=6\), and \(g(3)=8\); not continuous.

- Write the polynomial in factored form using the given zeros and their multiplicities. - At a continuous joining point, the two pieces must approach the same value. - Determine the value the polynomial must have at \(x=2\). - Substitute the joining point to find the scale factor.

1. A cubic polynomial with a double zero at \(0\) and a simple zero at \(4\) has the form \(p(x)=ax^2(x-4)\). 2. Continuity at \(x=2\) requires \(p(2)=8\). 3. Substitute \(x=2\): \(a\cdot 2^2\cdot(2-4)=8\), so \(-8a=8\). 4. Therefore, \(a=-1\), and \(p(x)=-x^2(x-4)=-x^3+4x^2\).

\(p(x)=-x^2(x-4)\), or equivalently \(p(x)=-x^3+4x^2\)

- Find the value that the left-hand piece approaches at the joining point. - Write a general linear function. - Translate the given zero into an equation. - Use the two conditions to form a system of equations.

1. The left-hand limit at \(x=-2\) is \(0.5\cdot(-2)^2+2\cdot(-2)+1=2-4+1=-1\). 2. Continuity requires \(g(-2)=-1\). 3. Write \(g(x)=mx+b\). The zero at \(x=3\) gives \(3m+b=0\), and continuity gives \(-2m+b=-1\). 4. Subtracting the equations gives \(5m=1\), so \(m=0.2\). Then \(b=-3m=-0.6\). 5. Therefore, \(g(x)=0.2x-0.6\).

\(g(x)=0.2x-0.6\)

- Trace each graph toward \(x=1\) from the left and from the right. - A filled point gives the function value; an open circle is not included. - For continuity, the left-hand limit, right-hand limit, and function value must agree. - Analyze the two panels separately.

1. In panel a), \(\lim_{x\to1^-}f(x)=1\), \(\lim_{x\to1^+}f(x)=-1\), and the filled point shows that \(f(1)=1\). Since the one-sided limits are different, the two-sided limit does not exist. Therefore, \(f\) has a jump discontinuity at \(x=1\). 2. In panel b), the left-hand piece gives \(\lim_{x\to1^-}g(x)=1^2-1=0\), and the right-hand piece gives \(\lim_{x\to1^+}g(x)=-1+1=0\). The pieces meet at \((1, 0)\), so \(g(1)=0\). The two one-sided limits and the function value are equal, so \(g\) is continuous at \(x=1\).

a) \(f\) is not continuous at \(x=1\): \(\lim_{x\to1^-}f(x)=1\), \(\lim_{x\to1^+}f(x)=-1\), and \(f(1)=1\). b) \(g\) is continuous at \(x=1\) because \(\lim_{x\to1^-}g(x)=\lim_{x\to1^+}g(x)=g(1)=0\).

- Trace the graph toward each input from both directions. - Compare the left-hand and right-hand limits. - Then compare the common limit, when it exists, with the function value. - Use the filled point at \(x=1\) to identify \(f(1)\).

1. At \(x=-1\), the left-hand piece approaches \(-1\), and the middle piece also approaches \(-1\). The graph includes the point \((-1, -1)\), so \(f(-1)=-1\). Thus \(\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)=-1\), and \(f\) is continuous at \(x=-1\). 2. At \(x=1\), the middle piece approaches \(-1\) from the left, while the right-hand piece approaches \(1\). The filled point gives \(f(1)=1\). Since \(\lim_{x\to1^-}f(x)=-1\ne1=\lim_{x\to1^+}f(x)\), the two-sided limit does not exist. Therefore, \(f\) is not continuous at \(x=1\).

At \(x=-1\): \(\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)=-1\), so \(f\) is continuous. At \(x=1\): \(\lim_{x\to1^-}f(x)=-1\), \(\lim_{x\to1^+}f(x)=1\), and \(f(1)=1\), so \(f\) is not continuous.

- Find the value reached by the first piece at the joining point. - Determine the value the second piece approaches as \(x\) approaches \(2\) from the right. - Set those values equal and solve for the parameter.

1. The first piece gives \(f(2)=2^2-3=1\). 2. The right-hand limit is \(\lim_{x\to2^+}(ax-1)=2a-1\). 3. Continuity at \(x=2\) requires \(2a-1=1\). 4. Solving gives \(2a=2\), so \(a=1\).

\(a=1\)

- Compute the left-hand expression at the boundary using the parameter. - Use the second branch for both the right-hand limit and the function value at \(2\). - Set the two one-sided limits equal and solve for \(a\).

1. The left-hand limit at \(x=2\) is \(2a+1\). 2. The right-hand limit and function value are \(F_a(2)=2^2-a=4-a\). 3. Continuity requires \(2a+1=4-a\), so \(3a=3\) and \(a=1\). Then \(F_a(2)=3\).

\(a=1\), and then \(F_a(2)=3\).

- Start with the nonnegative-radicand condition. - Use the numerator zero and denominator zero as sign-analysis breakpoints. - Apply continuity of a composition only on inputs where the inner function is defined and nonnegative.

1. The radicand must satisfy \(\frac{x-1}{x+2}\ge0\), with \(x\ne-2\). 2. A sign analysis at the critical values \(-2\) and \(1\) gives \(x<-2\) or \(x\ge1\). 3. On this domain the rational inner function is continuous and nonnegative, and square root is continuous on nonnegative inputs. Thus \(F\) is continuous on \((-\infty, -2)\) and \([1, \infty)\).

Domain: \((-\infty, -2)\cup[1, \infty)\). Continuity intervals: \((-\infty, -2)\) and \([1, \infty)\).

- Represent the double zero with a squared factor. - Find the value \(g(-2)\) must have for continuity. - Use a factored cubic so the known zero is built into the expression. - Apply the point condition and the joining-point condition to determine the remaining coefficients.

1. The left-hand piece approaches \(-2+5=3\) at \(x=-2\), so continuity requires \(g(-2)=3\). 2. A cubic with a double zero at \(x=1\) can be written as \(g(x)=(ax+b)(x-1)^2\). 3. Since \(g(0)=2\), \(b\cdot(0-1)^2=2\), so \(b=2\). 4. Use \(g(-2)=3\): \((-2a+2)(-3)^2=3\). Thus, \(9(-2a+2)=3\), so \(-18a=-15\) and \(a=\frac{5}{6}\). 5. Therefore, \(g(x)=\left(\frac{5}{6}x+2\right)(x-1)^2\). 6. Expanding gives \(g(x)=\frac{5}{6}x^3+\frac{1}{3}x^2-\frac{19}{6}x+2\).

\(g(x)=\left(\frac{5}{6}x+2\right)(x-1)^2\), or equivalently \(g(x)=\frac{5}{6}x^3+\frac{1}{3}x^2-\frac{19}{6}x+2\)

- Factor the numerator using the denominator factor \(x-3\). - Continuity requires the point value to equal the two-sided limit. - A vertical asymptote requires unbounded behavior, not merely a mismatched point value.

1. Factor the numerator: \(x^2+(k-3)x-3k=(x-3)(x+k)\). Thus, for \(x\ne3\), \(F(x)=x+k\). 2. Therefore \(\lim_{x\to3}F(x)=3+k\). Setting \(3+k=7\) gives \(k=4\). 3. With \(k=4\), continuity at \(x=3\) requires \(m=\lim_{x\to3}F(x)=7\). 4. If \(m=2\), the finite limit still equals \(7\) but differs from the point value. The discontinuity is removable, and \(x=3\) is not a vertical asymptote.

a) \(k=4\) b) \(m=7\) c) The discontinuity is removable, and \(x=3\) is not a vertical asymptote.