Free Math Worksheets

- What derivative relationship defines an antiderivative? - Use the power rule for differentiation. - What happens to an additive constant when you differentiate? - Match coefficients and exponents after differentiating \(F\).

1. Because \(F\) is an antiderivative of \(f\), \(F'(x)=f(x)\). 2. For part a, \(F'(x)=kx^{k-1}\). Matching \(7x^6\) gives \(k=7\). 3. For part b, \(F'(x)=8x^3\). Matching \(kx^3\) gives \(k=8\). 4. For part c, \(F'(x)=\cos(x)\) for every real \(k\), because \(k^2\) is constant. Thus, any real value of \(k\) works.

a) \(k=7\) b) \(k=8\) c) Any \(k\in\mathbb{R}\), such as \(k=0\)

- What derivative relationship defines an antiderivative? - Which differentiation rule applies to a function inside another function? - Account for the derivative of the inner expression. - Differentiate before comparing the functions.

1. For part a, differentiate using the chain rule: \(F'(x)=\frac{1}{x^2+5}(2x)=\frac{2x}{x^2+5}=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 2. For part b, \(F'(x)=\cos(x^2)(2x)=2x\cos(x^2)=f(x)\). Therefore, \(F\) is an antiderivative of \(f\).

a) Yes, \(F\) is an antiderivative of \(f\). b) Yes, \(F\) is an antiderivative of \(f\).

- What derivative relationship defines an antiderivative? - Differentiate \(F\). - Which rule applies to a square root of a polynomial?

1. Rewrite \(F(x)=(2x^2+3)^{\frac{1}{2}}\). 2. Differentiate using the chain rule: \(F'(x)=\frac{1}{2}(2x^2+3)^{-\frac{1}{2}}(4x)\). 3. Simplify: \(F'(x)=\frac{2x}{\sqrt{2x^2+3}}=f(x)\). 4. Therefore, \(F\) is an antiderivative of \(f\).

Yes. Since \(F^{\prime}(x)=\frac{2x}{\sqrt{2x^2+3}}=f(x)\), \(F\) is an antiderivative of \(f\).

- Recall the derivative relationship between a function and its antiderivative. - Which rule is needed to differentiate a product? - Compare the entire derivative with \(g\).

1. Differentiate \(G(x)=x\sin(x)\) using the product rule. 2. \(G'(x)=\sin(x)+x\cos(x)\). 3. Since \(\sin(x)+x\cos(x)\ne\cos(x)\), \(G\) is not an antiderivative of \(g\).

No. \(G^{\prime}(x)=\sin(x)+x\cos(x)\ne g(x)\).

- Integrate each term separately. - Recall an antiderivative of \(\cos(x)\). - Different constants of integration produce different antiderivatives.

1. A general antiderivative is \(F(x)=6\sin(x)-2x^2+C\). 2. Choose two different constants, such as \(C=0\) and \(C=5\). This gives \(F_1(x)=6\sin(x)-2x^2\) and \(F_2(x)=6\sin(x)-2x^2+5\).

\(F_1(x)=6\sin(x)-2x^2\) and \(F_2(x)=6\sin(x)-2x^2+5\)

- What equation must hold for \(f\) to be an antiderivative of \(g\)? - Use the product rule on the nonconstant term. - What happens to the constant \(5\) when differentiated? - Factor out the common exponential term after differentiating.

1. Differentiate the product \((x^3-4x)e^x\) using the product rule. The constant \(5\) differentiates to \(0\). 2. \(f'(x)=(3x^2-4)e^x+(x^3-4x)e^x\). 3. Factor out \(e^x\): \(f'(x)=(x^3+3x^2-4x-4)e^x\). 4. This equals \(g(x)\), so \(f\) is an antiderivative of \(g\).

\(f^{\prime}(x)=(3x^2-4)e^x+(x^3-4x)e^x=(x^3+3x^2-4x-4)e^x=g(x)\). Therefore, \(f\) is an antiderivative of \(g\).

- What derivative relationship defines an antiderivative? - Apply the chain rule to the linear inner expressions. - Track the sign when differentiating cosine. - Constant terms disappear when differentiated.

1. For part a, differentiate using the chain rule: \(F'(x)=4e^{\frac{1}{2}x+1}\left(\frac{1}{2}\right)=2e^{\frac{1}{2}x+1}=f(x)\). Thus, \(F\) is an antiderivative of \(f\). 2. For part b, \(F'(x)=\frac{1}{2}[-\sin(2x)](2)=-\sin(2x)\). Since this is not \(f(x)=\sin(2x)\), \(F\) is not an antiderivative of \(f\).

a) Yes, \(F\) is an antiderivative of \(f\). b) No, \(F\) is not an antiderivative of \(f\).

- Work term by term using the sum rule. - Use the basic derivative and antiderivative rules for powers, exponential functions, sine, and cosine. - Check the signs of the trigonometric terms. - Include an arbitrary constant in the general antiderivative.

1. Differentiate term by term: \(f'(x)=4e^x-2x^3+\cos x\). 2. Differentiate again: \(f''(x)=4e^x-6x^2-\sin x\). 3. Integrate the original function term by term: \(F(x)=4e^x-\frac{1}{10}x^5-\cos x+C\). Differentiating this expression returns \(f(x)\).

\(f'(x)=4e^x-2x^3+\cos x\) \(f''(x)=4e^x-6x^2-\sin x\) \(F(x)=4e^x-\frac{1}{10}x^5-\cos x+C\)

- What equation must hold for \(F\) to be an antiderivative of \(f\)? - Use the chain rule. - Identify the inner and outer functions. - What is the derivative of cosine?

1. Differentiate \(F\) using the chain rule. 2. The derivative of the outer function \(2\cos(u)\) is \(-2\sin(u)\), and the derivative of the inner function \(u=x^3\) is \(3x^2\). 3. Thus, \(F'(x)=-2\sin(x^3)(3x^2)=-6x^2\sin(x^3)=f(x)\). 4. Therefore, \(F\) is an antiderivative of \(f\).

\(F^{\prime}(x)=-6x^2\sin(x^3)=f(x)\), so \(F\) is an antiderivative of \(f\).

- Differentiate each candidate and compare it with \(f\). - Use the product rule for a polynomial times \(e^x\). - What happens to an added constant when differentiated? - Compare the polynomial coefficients after factoring out \(e^x\).

1. Differentiate each product using the product rule. 2. \(F_A'(x)=(x^2+7x+10)e^x\ne f(x)\). 3. \(F_B'(x)=(x^2+6x+5)e^x\ne f(x)\). 4. \(F_C'(x)=(x^2+5x+5)e^x=f(x)\). The constant \(4\) differentiates to \(0\). 5. \(F_D'(x)=(x^2+9x+17)e^x\ne f(x)\). 6. Therefore, \(F_C\) is an antiderivative of \(f\).

\(F_C(x)=(x^2+3x+2)e^x+4\)

- What equation must hold for \(F\) to be an antiderivative of \(f\)? - Split the fraction into separate powers before differentiating. - Apply the power rule to negative exponents.

1. Simplify \(F\): \(F(x)=2x^2-3x^{-2}\). 2. Differentiate using the power rule: \(F'(x)=4x+6x^{-3}\). 3. Since \(F'(x)=f(x)\), \(F\) is an antiderivative of \(f\).

Yes, \(F\) is an antiderivative of \(f\).

- Treat \(\frac1a\) as a constant factor. - Rewrite the variable in the denominator using a negative exponent. - Differentiate the given antiderivative and compare coefficients.

1. Rewrite \(h_a(x)=\frac1a x^{-2}\). Integrating gives \(H_a(x)=-\frac{1}{ax}+C\). 2. Differentiate the given function: \(H^{\prime}(x)=\frac{1}{5x^2}\). Comparing this with \(h_a(x)=\frac{1}{ax^2}\) gives \(a=5\).

1. \(H_a(x)=-\frac{1}{ax}+C\) 2. \(a=5\)

- Rewrite quotients using negative exponents. - Apply the power rule carefully to the exponent and coefficient. - Differentiate each proposed antiderivative to verify the match.

1. \(\int12(x+2)^{-4}\,dx=-4(x+2)^{-3}\), so A matches 1. 2. \(\int12x^{-4}\,dx=-4x^{-3}\), so B matches 2. 3. \(\int-12x^{-5}\,dx=3x^{-4}\), so C matches 3. 4. \(\int12(x+2)^{-5}\,dx=-3(x+2)^{-4}\), so D matches 4.

A–1 B–2 C–3 D–4

- What derivative relationship defines an antiderivative? - Differentiate the sum term by term. - What is the derivative of \(\ln|x|\)? - Keep the constant factors when differentiating.

1. Differentiate term by term. 2. \(\frac{d}{dx}\left(\frac{1}{3}x^3\right)=x^2\). 3. For \(x\ne0\), \(\frac{d}{dx}[2\ln|x|]=\frac{2}{x}\). 4. Thus, \(F'(x)=x^2+\frac{2}{x}=f(x)\), so \(F\) is an antiderivative of \(f\).

\(F^{\prime}(x)=x^2+\frac{2}{x}=f(x)\), so \(F\) is an antiderivative of \(f\).

- Use the derivative definition of an antiderivative. - Apply the chain rule to the logarithm. - Differentiate the expression inside the logarithm separately. - Recall that \(\frac{d}{dx}\ln(u)=\frac{u'}{u}\).

1. Differentiate \(F(x)=\ln(u(x))\) with \(u(x)=x^2+e^x\). 2. Since \(u'(x)=2x+e^x\), the chain rule gives \(F'(x)=\frac{u'(x)}{u(x)}\). 3. Therefore, \(F'(x)=\frac{2x+e^x}{x^2+e^x}=f(x)\). 4. Hence, \(F\) is an antiderivative of \(f\).

\(F^{\prime}(x)=\frac{2x+e^x}{x^2+e^x}=f(x)\), so \(F\) is an antiderivative of \(f\).

- Recall an antiderivative of \(\frac1x\) on \(x>0\). - Integrate each term separately. - Different constants produce different antiderivatives.

1. A general antiderivative is \(F(x)=6\ln(x)+2x+C\). 2. Choose two different constants, such as \(C=0\) and \(C=5\). Then \(F_1(x)=6\ln(x)+2x\) and \(F_2(x)=6\ln(x)+2x+5\).

\(F_1(x)=6\ln(x)+2x\) and \(F_2(x)=6\ln(x)+2x+5\)

- Reverse the power rule by increasing the exponent by \(1\) and dividing by the new exponent. - Choose each coefficient so that differentiating reproduces the given coefficient. - Differentiate each result to check it.

1. For a), reverse the power rule: \(f(x)=4x^3\). Check: \(f^{\prime}(x)=12x^2\). 2. For b), \(f(x)=\frac25x^5\). Check: \(f^{\prime}(x)=2x^4\). 3. For c), \(f(x)=\frac15x^4\). Check: \(f^{\prime}(x)=\frac45x^3\). 4. For d), \(f(x)=10x\). Check: \(f^{\prime}(x)=10\).

a) \(f(x)=4x^3\) b) \(f(x)=\frac25x^5\) c) \(f(x)=\frac15x^4\) d) \(f(x)=10x\)

- Apply the reverse power rule to each term. - A constant derivative comes from a linear term. - Differentiate your result to check it.

1. Integrate each term using the reverse power rule: \(12x^3\) gives \(3x^4\), \(-5x^4\) gives \(-x^5\), and \(2\) gives \(2x\). 2. Thus one possible function is \(f(x)=3x^4-x^5+2x\). 3. Differentiating gives \(f^{\prime}(x)=12x^3-5x^4+2\), as required.

\(f(x)=3x^4-x^5+2x\). More generally, \(f(x)=3x^4-x^5+2x+C\), where \(C\in\mathbb{R}\).

- Use the power rule in reverse to find the general antiderivative. - A point on the graph gives an equation involving \(C\). - Substitute each point separately.

1. The general antiderivative is \(F(x)=2x^3-2x^2+x+C\). 2. For part a, \(F(0)=7\) gives \(C=7\). 3. For part b, \(F(1)=2\) gives \(1+C=2\), so \(C=1\). 4. For part c, \(F(-1)=0\) gives \(-2-2-1+C=0\), so \(C=5\).

a) \(F(x)=2x^3-2x^2+x+7\) b) \(F(x)=2x^3-2x^2+x+1\) c) \(F(x)=2x^3-2x^2+x+5\)

- First find the general antiderivative. - All antiderivatives of the same function differ by a constant. - Choose three different values for that constant.

1. Integrating term by term gives the general antiderivative \(F(x)=2x^4-2x^3+5x+C\). 2. Choose three different constants, such as \(C=0\), \(C=3\), and \(C=-1\). 3. This gives \(F_1(x)=2x^4-2x^3+5x\), \(F_2(x)=2x^4-2x^3+5x+3\), and \(F_3(x)=2x^4-2x^3+5x-1\).

\(F_1(x)=2x^4-2x^3+5x\) \(F_2(x)=2x^4-2x^3+5x+3\) \(F_3(x)=2x^4-2x^3+5x-1\)

- Expand the squared binomial before integrating. - Integrate each term of the resulting polynomial separately. - Use the reverse power rule.

1. Expand the square: \(f(x)=\frac12x^4-3x^2+\frac92\). 2. Integrate term by term to obtain \(F(x)=\frac1{10}x^5-x^3+\frac92x\). 3. Differentiating this expression reproduces \(f(x)\).

\(F(x)=\frac1{10}x^5-x^3+\frac92x\)

- Differentiate the proposed antiderivative. - What must the derivative of an antiderivative equal? - What effect does an additive constant have on a derivative?

1. Differentiate \(F\): \(F'(x)=4x^3-6x=f(x)\). Thus, \(F\) is an antiderivative of \(f\). 2. A function is an antiderivative of \(f\) on an interval when its derivative equals \(f\) at every point of the interval. 3. Differentiate \(H\): \(H'(x)=4x^3-6x=f(x)\). The different constant term has no effect on the derivative. 4. Therefore, \(H\) is also an antiderivative of \(f\).

a) Yes, because \(F'(x)=4x^3-6x=f(x)\). b) An antiderivative has derivative \(f\). Since \(H'(x)=4x^3-6x=f(x)\), \(H\) is also an antiderivative.

- Use the power rule for antiderivatives. - What effect does an additive constant have on the derivative? - A point on the graph gives an equation for \(C\). - Substitute \(x=1\) and the given output.

1. Rewrite \(f(x)=x^{-2}+1\). The family of antiderivatives is \(F(x)=-\frac{1}{x}+x+C\). 2. Four examples are obtained by choosing different constants, such as \(C=0,1,-5,10\). 3. For part b, use \(F(1)=0\): \(-1+1+C=0\), so \(C=0\). 4. Therefore, the required antiderivative is \(F(x)=-\frac{1}{x}+x\).

a) Examples: \(F_1(x)=-\frac{1}{x}+x\), \(F_2(x)=-\frac{1}{x}+x+1\), \(F_3(x)=-\frac{1}{x}+x-5\), and \(F_4(x)=-\frac{1}{x}+x+10\) b) \(F(x)=-\frac{1}{x}+x\)

- An antiderivative of a linear function is quadratic. - Use the zero of \(f\) to locate a horizontal tangent of \(F\). - Use \(F(2)=0\) to determine the vertical position of the graph.

1. The graph is the line \(f(x)=\frac12x-1\). 2. An antiderivative is \(F(x)=\frac14x^2-x+C\). 3. Use \(F(2)=0\): \(1-2+C=0\), so \(C=1\). 4. Therefore, \(F(x)=\frac14x^2-x+1=\frac14(x-2)^2\). Its graph is an upward-opening parabola with vertex \((2,0)\).

\(F(x)=\frac14(x-2)^2\), an upward-opening parabola with vertex \((2,0)\)

- Use the chain rule to differentiate the candidate antiderivatives. - Look for a numerator that is the derivative of the denominator. - For a linear inner function, account for its constant derivative. - Differentiate each candidate to check your matches.

1. Differentiate \(F_C\): \(F_C'(x)=4\cos(4x-2)=f_1(x)\). Thus \(f_1\) matches \(F_C\). 2. Since the derivative of \(x^2+x\) is \(2x+1\), \(\frac{d}{dx}\ln|x^2+x|=\frac{2x+1}{x^2+x}=f_2(x)\). Thus \(f_2\) matches \(F_A\). 3. Differentiate \(F_D\): \(F_D'(x)=\frac{1}{10}\cdot 5(2x-3)^4\cdot 2=(2x-3)^4=f_3(x)\). Thus \(f_3\) matches \(F_D\). 4. Differentiate \(F_B\): \(F_B'(x)=2\cdot 0.5e^{0.5x}=e^{0.5x}=f_4(x)\). Thus \(f_4\) matches \(F_B\).

1. C 2. A 3. D 4. B

- Identify the function variable in each part. - Treat all other symbols as constants. - Rewrite a variable in the denominator using a negative exponent before applying the power rule. - Differentiate your result to check it.

1. Integrate with respect to \(x\): \(F(x)=\frac{4}{k}\cdot\frac{x^4}{4}-3a\cdot\frac{x^3}{3}=\frac{x^4}{k}-ax^3\). 2. Rewrite \(\frac{k}{t^2}=kt^{-2}\) and integrate with respect to \(t\): \(G(t)=-kt^{-1}+at=-\frac{k}{t}+at\). 3. Integrate with respect to \(a\), treating \(x\) as a constant: \(H(a)=\frac1x\cdot\frac{a^2}{2}+a\sqrt{x}=\frac{a^2}{2x}+a\sqrt{x}\).

a) \(F(x)=\frac{x^4}{k}-ax^3\) b) \(G(t)=-\frac{k}{t}+at\) c) \(H(a)=\frac{a^2}{2x}+a\sqrt{x}\)

- Check the sign when integrating \(-\sin(x)\). - The exponential function \(e^t\) is its own antiderivative. - Apply the power rule to fractional exponents. - Differentiate each result to verify it.

1. Integrate with respect to \(x\): \(F(x)=\frac1k\cos(x)+k^2x\). 2. Integrate with respect to \(t\): \(G(t)=e^t-\frac12k^2t^2\). 3. Apply the power rule with respect to \(z\): \(H(z)=\frac25z^{5/2}+\frac52a^2z\).

a) \(F(x)=\frac1k\cos(x)+k^2x\) b) \(G(t)=e^t-\frac12k^2t^2\) c) \(H(z)=\frac25z^{5/2}+\frac52a^2z\)

- Which integration rules apply to the two terms? - What equation follows from the point on the graph? - How can you use that equation to find \(C\)? - Keep the logarithm's domain in mind.

1. Integrate each term: \(F(x)=\frac{1}{2}e^{2x}+\ln(x+2)+C\). The domain condition \(x>-2\) makes \(x+2>0\). 2. Use \(F(0)=1.5\): \(\frac{1}{2}+\ln(2)+C=1.5\). 3. Solve for the constant: \(C=1-\ln(2)\). 4. Therefore, \(F(x)=\frac{1}{2}e^{2x}+\ln(x+2)+1-\ln(2)\).

\(F(x)=\frac{1}{2}e^{2x}+\ln(x+2)+1-\ln(2)\)

- Find the general antiderivative of \(\cos(kx)\). - Determine the range of \(\frac{1}{k}\sin(kx)\). - How does adding \(C\) shift the range vertically? - Set the minimum of the shifted range equal to \(0\).

1. The general antiderivative is \(F_k(x)=\frac{1}{k}\sin(kx)+C\). 2. Since \(\sin(kx)\) ranges from \(-1\) to \(1\), the term \(\frac{1}{k}\sin(kx)\) ranges from \(-\frac{1}{|k|}\) to \(\frac{1}{|k|}\). 3. Therefore, the minimum value of \(F_k\) is \(C-\frac{1}{|k|}\). 4. Set the minimum equal to \(0\): \(C-\frac{1}{|k|}=0\), so \(C=\frac{1}{|k|}\). 5. Thus \(F_k(x)=\frac{1}{k}\sin(kx)+\frac{1}{|k|}\).

\(F_k(x)=\frac{1}{k}\sin(kx)+\frac{1}{|k|}\)

- First find the general antiderivative of \(g\). - Determine the range of the cosine term. - What must be true of the minimum and maximum for the function to have both signs? - Solve the resulting inequalities for \(C\).

1. The general antiderivative is \(G(x)=-2\cos(3x+2)+C\). 2. Because cosine ranges from \(-1\) to \(1\), the term \(-2\cos(3x+2)\) ranges from \(-2\) to \(2\). Thus the range of \(G\) is \([C-2, C+2]\). 3. For \(G\) to take both positive and negative values, its minimum must be below \(0\) and its maximum above \(0\): \(C-2<0\) and \(C+2>0\). 4. These inequalities give \(-2<C<2\).

\(G(x)=-2\cos(3x+2)+C\), where \(C\in(-2, 2)\)

- Find an antiderivative of each term separately. - Check the sign that comes from the derivative of \(2-x\). - Use the given value to solve for the integration constant. - Differentiate your result to verify it.

1. Integrate each term: \(F(x)=\frac{1}{8}x^4+e^{2-x}+C\). 2. Apply \(F(2)=10\): \(\frac{1}{8}(2)^4+e^0+C=10\). 3. Simplify: \(2+1+C=10\), so \(C=7\). 4. Therefore, \(F(x)=\frac{1}{8}x^4+e^{2-x}+7\).

\(F(x)=\frac{1}{8}x^4+e^{2-x}+7\)

- Which function has a derivative of the form \(\frac{1}{x+a}\)? - Account for the linear inner function when integrating the cosine term. - Use \(F(0)=0\) to isolate \(C\). - A logarithm property can combine the two logarithmic terms.

1. Integrate each term: \(F(x)=2\ln(x+3)-2\sin\left(\frac{x}{2}\right)+C\). The domain condition ensures \(x+3>0\). 2. Apply \(F(0)=0\): \(2\ln(3)-2\sin(0)+C=0\). 3. Therefore, \(C=-2\ln(3)\). 4. The required antiderivative is \(F(x)=2\ln(x+3)-2\sin\left(\frac{x}{2}\right)-2\ln(3)\).

\(F(x)=2\ln(x+3)-2\sin\left(\frac{x}{2}\right)-2\ln(3)\), equivalently \(F(x)=2\ln\left(\frac{x+3}{3}\right)-2\sin\left(\frac{x}{2}\right)\)

- Integrate the exponential and polynomial terms separately. - How does the derivative of \(3x+3\) affect the exponential antiderivative? - Substitute the given input and output to find \(C\). - Solve the resulting equation for the unknown constant.

1. Integrate term by term: \(F(x)=\frac{2}{3}e^{3x+3}-2x^2+C\). 2. Apply \(F(-1)=5\): \(\frac{2}{3}e^0-2(-1)^2+C=5\). 3. Simplify: \(\frac{2}{3}-2+C=5\), so \(-\frac{4}{3}+C=5\). 4. Thus, \(C=\frac{19}{3}\), and \(F(x)=\frac{2}{3}e^{3x+3}-2x^2+\frac{19}{3}\).

\(F(x)=\frac{2}{3}e^{3x+3}-2x^2+\frac{19}{3}\)

- How do you differentiate a composite function? - What equation must hold for one function to be an antiderivative of another? - Compose the functions before integrating. - How does a specified function value determine the integration constant?

1. For part a, compose the functions: \(h(x)=f(g(x))=\left(\frac{1}{3}x+2\right)^2-5\). 2. Differentiate \(H\): \(H'(x)=3\left(\frac{1}{3}x+2\right)^2\left(\frac{1}{3}\right)-5=\left(\frac{1}{3}x+2\right)^2-5=h(x)\). Therefore, \(H\) is an antiderivative of \(h\). 3. For part b, \(k(x)=6g(f(x))=6\left(\frac{1}{3}(x^2-5)+2\right)=2x^2+2\). 4. The general antiderivative is \(K(x)=\frac{2}{3}x^3+2x+C\). 5. Use \(K(3)=20\): \(\frac{2}{3}(3)^3+2(3)+C=20\), so \(24+C=20\) and \(C=-4\). 6. Thus, \(K(x)=\frac{2}{3}x^3+2x-4\).

a) \(H'(x)=\left(\frac{1}{3}x+2\right)^2-5=f(g(x))\), so \(H\) is an antiderivative of \(h\). b) \(K(x)=\frac{2}{3}x^3+2x-4\)

- Use the condition \(F'(x)=f(x)\). - Pay close attention to the chain rule in part c. - In part b, compare exponents first. - Also verify that the coefficients match.

1. In each part, require \(F'(x)=f(x)\). 2. For part a, \(F'(x)=3ax^2\). Matching coefficients with \(\frac{1}{3}x^2\) gives \(3a=\frac{1}{3}\), so \(a=\frac{1}{9}\). 3. For part b, \(F'(x)=4x^3\). Matching exponents gives \(a-2=3\), so \(a=5\). Then \(a-1=4\), so the coefficient also matches. 4. For part c, \(F'(x)=3ae^{3x}\). Matching coefficients with \(e^{3x}\) gives \(3a=1\), so \(a=\frac{1}{3}\).

a) \(a=\frac{1}{9}\) b) \(a=5\) c) \(a=\frac{1}{3}\)

- Which differentiation rule applies to the product in part a? - Recall the product rule. - Rewrite the fraction in part b using a negative exponent. - Track the negative sign and the derivative of the inner expression.

1. For part a, use the product rule: \(F'(x)=e^x+(x-1)e^x=xe^x=f(x)\). The statement is true. 2. For part b, write \(F(x)=(x^2+1)^{-1}\). Then \(F'(x)=-(x^2+1)^{-2}(2x)=-\frac{2x}{(x^2+1)^2}\), which is not equal to \(f(x)\). The statement is false.

a) True, because \(F'(x)=xe^x=f(x)\). b) False, because \(F'(x)=-\frac{2x}{(x^2+1)^2}\ne f(x)\).

- Differentiate each proposed antiderivative. - Use the power rule in reverse for functions that have no match. - An antiderivative \(F\) must satisfy \(F'=f\). - You may choose any additive constant when none is specified.

1. Differentiate the listed antiderivatives: \(F_1'(x)=2x+3\), \(F_2'(x)=6x^2\), \(F_3'(x)=4x^3-x\), and \(F_4'(x)=5x^4\). 2. Therefore, the matches are A-2, B-1, D-3, and E-4. 3. For C, one antiderivative is \(\int \frac{1}{2}x^3\,dx=\frac{1}{8}x^4\). 4. For F, one antiderivative is \(\int 3\,dx=3x\).

Matches: A-2, B-1, D-3, E-4. For C: \(F(x)=\frac{1}{8}x^4\) For F: \(F(x)=3x\)

- How are derivatives and antiderivatives related? - First write each general antiderivative with a constant \(C\). - Substitute the coordinates of the given point to determine \(C\).

1. For part a, \(G_A'(x)=4x-1=g_1(x)\), and \(G_B'(x)=2x^3-\frac{1}{2}x^2=g_4(x)\). 2. For \(g_2(x)=3x^2-4x\), the general antiderivative is \(G_2(x)=x^3-2x^2+C\). Using \(G_2(1)=2\) gives \(-1+C=2\), so \(C=3\). Thus, \(G_2(x)=x^3-2x^2+3\). 3. For \(g_3(x)=x^3\), the general antiderivative is \(G_3(x)=\frac{1}{4}x^4+C\). Using \(G_3(1)=2\) gives \(\frac{1}{4}+C=2\), so \(C=\frac{7}{4}\). Thus, \(G_3(x)=\frac{1}{4}x^4+\frac{7}{4}\).

a) \(G_A\) matches \(g_1\), and \(G_B\) matches \(g_4\). b) \(G_2(x)=x^3-2x^2+3\) \(G_3(x)=\frac{1}{4}x^4+\frac{7}{4}\)

- How is the slope of \(F\) related to \(f\)? - When are two tangent lines parallel? - What equation gives all points with the same slope? - Use the given point to determine \(C\).

1. The slope of \(F\) at \(x=1\) is \(F'(1)=f(1)=1-4+5=2\). 2. Parallel tangent lines have equal slopes, so solve \(f(x)=2\): \(x^2-4x+5=2\). 3. This becomes \(x^2-4x+3=0\), or \((x-1)(x-3)=0\). Besides \(x=1\), the other solution is \(x=3\). 4. For part c, the general antiderivative is \(G(x)=\frac{1}{3}x^3-2x^2+5x+C\). 5. Apply \(G(3)=8\): \(9-18+15+C=8\), so \(6+C=8\) and \(C=2\). 6. Therefore, \(G(x)=\frac{1}{3}x^3-2x^2+5x+2\).

a) \(2\) b) Yes. The other point has x-coordinate \(x=3\). c) \(G(x)=\frac{1}{3}x^3-2x^2+5x+2\)

- Split the fraction into separate terms. - Check the sign when integrating \(\sin(x)\). - Use the given point to determine the constant of integration.

1. Rewrite \(g(x)=\frac35\sin(x)-\frac25\). A general antiderivative is \(G(x)=-\frac35\cos(x)-\frac25x+C\). 2. For part a, choose \(C=0\): \(G_1(x)=-\frac35\cos(x)-\frac25x\). 3. For part b, use \(G_2(0)=1\): \(-\frac35+C=1\), so \(C=\frac85\). Therefore, \(G_2(x)=-\frac35\cos(x)-\frac25x+\frac85\).

a) \(G_1(x)=-\frac35\cos(x)-\frac25x\) b) \(G_2(x)=-\frac35\cos(x)-\frac25x+\frac85\)

- What derivative relationship defines an antiderivative? - Apply the product rule separately to the two products. - Track the signs of the trigonometric derivatives carefully. - Look for terms that cancel after differentiation.

1. Differentiate both terms of \(f\) using the product rule. 2. The derivative of \((x^2-2)\cos(x)\) is \(2x\cos(x)-(x^2-2)\sin(x)\). 3. The derivative of \(-2x\sin(x)\) is \(-2\sin(x)-2x\cos(x)\). 4. Combine the results: \(f'(x)=2x\cos(x)-(x^2-2)\sin(x)-2\sin(x)-2x\cos(x)\). 5. The cosine terms cancel, and \(2\sin(x)-2\sin(x)=0\), leaving \(f'(x)=-x^2\sin(x)=g(x)\). 6. Therefore, \(f\) is an antiderivative of \(g\).

\(f^{\prime}(x)=2x\cos(x)-(x^2-2)\sin(x)-2\sin(x)-2x\cos(x)=-x^2\sin(x)=g(x)\). Therefore, \(f\) is an antiderivative of \(g\).

- Does the expression resemble the result of a familiar differentiation rule? - Look for two terms that could come from differentiating a product. - Find functions \(u\) and \(v\) whose product has derivative \(f\). - Differentiate your proposed antiderivative to check it.

1. Recognize the product-rule pattern \(u'v+uv'\). 2. Choose \(u(x)=x^3\) and \(v(x)=\sin(x)\). Then \(u'(x)=3x^2\) and \(v'(x)=\cos(x)\). 3. Therefore, \(\frac{d}{dx}[x^3\sin(x)]=3x^2\sin(x)+x^3\cos(x)=f(x)\). 4. One antiderivative is \(F(x)=x^3\sin(x)\).

\(F(x)=x^3\sin(x)\)

- Use the chain rule: derivative of the outer function times derivative of the inner function. - Rewrite a square root as a power when helpful. - Identify the inner expression in each part. - Compare the complete derivative with \(f\).

1. For part a, use the chain rule: \(F'(x)=5(x^2-4x)^4(2x-4)\). This equals \(f(x)\), so \(F\) is an antiderivative of \(f\). 2. For part b, write \(F(x)=(x^2+1)^{\frac{1}{2}}+4\). Then \(F'(x)=\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)=\frac{x}{\sqrt{x^2+1}}=f(x)\). Thus, \(F\) is an antiderivative of \(f\).

a) Yes, \(F\) is an antiderivative of \(f\). b) Yes, \(F\) is an antiderivative of \(f\).

- Track the signs when differentiating and integrating sine and cosine. - Treat \(\pi\) as a constant. - Apply the power rule term by term.

1. Differentiate term by term: \(f^{\prime}(x)=3x^3+\sin(x)\). 2. Integrate term by term: \(F(x)=\frac{3}{20}x^5-\sin(x)+\pi x\).

\(f^{\prime}(x)=3x^3+\sin(x)\) \(F(x)=\frac{3}{20}x^5-\sin(x)+\pi x\)

- Expand the squared binomial first. - Apply the derivative and antiderivative power rules term by term. - Differentiate the antiderivative to check it.

1. Expand the function: \(f(x)=x^2+4x+4-3x^3\). 2. Differentiate term by term: \(f^{\prime}(x)=-9x^2+2x+4\). 3. Integrate term by term: \(F(x)=-\frac34x^4+\frac13x^3+2x^2+4x\).

\(f^{\prime}(x)=-9x^2+2x+4\) \(F(x)=-\frac34x^4+\frac13x^3+2x^2+4x\)

- Differentiate and integrate each term separately. - Track the signs for sine and cosine. - Remember that \(e^x\) is its own derivative and antiderivative. - Use the first derivative to find slope at a point.

1. Differentiate term by term: \(g'(x)=-3\sin x-4e^x\). Differentiating again gives \(g''(x)=-3\cos x-4e^x\). 2. Integrate term by term. One antiderivative is \(G(x)=3\sin x-4e^x\). More generally, \(G(x)=3\sin x-4e^x+C\). 3. The slope at \(x=0\) is \(g'(0)=-3\sin 0-4e^0=-4\).

a) \(g'(x)=-3\sin x-4e^x\); \(g''(x)=-3\cos x-4e^x\) b) \(G(x)=3\sin x-4e^x+C\) c) The slope is \(-4\).

- Differentiate and integrate each term separately. - Include an arbitrary constant in the general antiderivative. - Substitute the coordinates of the given point into \(G\). - Use \(e^0=1\) to solve for the constant.

1. Differentiate term by term: \(g'(x)=4x-3e^x\) and \(g''(x)=4-3e^x\). 2. The general antiderivative is \(G(x)=\frac{2}{3}x^3-3e^x+x+C\). Use the condition \(G(0)=2\): \(-3+C=2\), so \(C=5\). Therefore, \(G(x)=\frac{2}{3}x^3-3e^x+x+5\).

a) \(g'(x)=4x-3e^x\); \(g''(x)=4-3e^x\) b) \(G(x)=\frac{2}{3}x^3-3e^x+x+5\)

- What derivative relationship defines an antiderivative? - Use the product rule. - Remember the chain-rule factor when differentiating \(e^{2x}\). - Factor out the common exponential term before comparing.

1. Use the product rule with \(u(x)=x^2-3x\) and \(v(x)=e^{2x}\). 2. Their derivatives are \(u'(x)=2x-3\) and \(v'(x)=2e^{2x}\). 3. Therefore, \(F'(x)=(2x-3)e^{2x}+2(x^2-3x)e^{2x}\). 4. Factor and simplify: \(F'(x)=e^{2x}(2x-3+2x^2-6x)=(2x^2-4x-3)e^{2x}=f(x)\). 5. Hence, \(F\) is an antiderivative of \(f\).

Yes. \(F^{\prime}(x)=(2x^2-4x-3)e^{2x}=f(x)\), so \(F\) is an antiderivative of \(f\).

- Use the zero-product property in part a. - Recall the quadratic formula. - Differentiate \(F\) to verify the antiderivative relationship. - Antiderivatives differ by a constant; use the initial value to find it.

1. For part a, \(e^x>0\), so the zeros satisfy \(x^2+4x+2=0\). 2. The quadratic formula gives \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\). 3. For part b, differentiate using the product rule: \(F'(x)=(2x+2)e^x+(x^2+2x)e^x=(x^2+4x+2)e^x=f(x)\). 4. Every antiderivative has the form \(G(x)=F(x)+C\). Since \(F(-2)=0\), the condition \(G(-2)=1\) gives \(C=1\). 5. Therefore, \(G(x)=(x^2+2x)e^x+1\).

a) \(x=-2-\sqrt{2}\) and \(x=-2+\sqrt{2}\) b) \(F'(x)=f(x)\), and \(G(x)=(x^2+2x)e^x+1\)

- Use \(F'(x)=f(x)\). - Differentiate with the product rule and factor out \(e^x\). - Equal polynomials have matching coefficients. - Solve the resulting three equations in order.

1. Differentiate \(F\) using the product rule: \(F'(x)=(2ax+b)e^x+(ax^2+bx+c)e^x\). 2. Combine terms: \(F'(x)=[ax^2+(2a+b)x+(b+c)]e^x\). 3. Match coefficients with \((x^2-x-2)e^x\): \(a=1\), \(2a+b=-1\), and \(b+c=-2\). 4. From \(a=1\), \(2+b=-1\), so \(b=-3\). Then \(-3+c=-2\), so \(c=1\). 5. Therefore, \(a=1\), \(b=-3\), and \(c=1\).

\(a=1\), \(b=-3\), \(c=1\)

- Which integration rules apply to the exponential, polynomial, and sine terms? - How does the coefficient in the exponent affect the antiderivative? - What equation does the point on the graph provide? - Use that equation to determine \(C\).

1. Integrate term by term: \(F(x)=2e^{2x}-\frac{1}{6}x^3-\cos(x)+C\). 2. Use \(F(0)=3\): \(2e^0-0-\cos(0)+C=3\). 3. This gives \(2-1+C=3\), so \(C=2\). 4. Therefore, \(F(x)=2e^{2x}-\frac{1}{6}x^3-\cos(x)+2\).

\(F(x)=2e^{2x}-\frac{1}{6}x^3-\cos(x)+2\)

- Differentiate \(F\) and compare the result with \(f\). - Expand and simplify \(F\) before differentiating. - Track signs carefully when differentiating negative powers. - Compare the numerators after using a common denominator.

1. Expand and simplify: \(F(x)=\frac{x^4-8x^2+16}{x}=x^3-8x+16x^{-1}\). 2. Differentiate: \(F'(x)=3x^2-8-16x^{-2}\). 3. With a common denominator, \(F'(x)=\frac{3x^4-8x^2-16}{x^2}\). 4. The constant term in the numerator is \(-16\), not \(+16\). Therefore, \(F'(x)\ne f(x)\), so \(F\) is not an antiderivative of \(f\).

No, \(F\) is not an antiderivative of \(f\).

- Rewrite the quotient using a negative exponent. - Apply the power rule for antiderivatives. - Different constants give different antiderivatives. - Substitute the given point to determine the constant.

1. Rewrite \(f_n(x)=3x^{-n}\). By the power rule, \(F_n(x)=\frac{3}{1-n}x^{1-n}+C=-\frac{3}{(n-1)x^{n-1}}+C\). 2. For each fixed \(n\), there are infinitely many antiderivatives because \(C\) can be any real number. 3. For \(n=4\), \(F_4(x)=-\frac1{x^3}+C\). Use \(F_4(1)=1\): \(-1+C=1\), so \(C=2\). Therefore, \(F_4(x)=-\frac1{x^3}+2\).

1. \(F_n(x)=-\frac{3}{(n-1)x^{n-1}}+C\), where \(C\in\mathbb R\) 2. Infinitely many 3. \(F_4(x)=-\frac1{x^3}+2\)

- The derivative of an antiderivative must equal the original function. - Use the quotient rule for rational functions with variable expressions in both numerator and denominator. - Rewrite reciprocal powers using negative exponents. - Simplify each derivative carefully before matching.

1. Differentiate \(F_1\) using the quotient rule: \(F_1'(x)=\frac{2(x+5)-2x}{(x+5)^2}=\frac{10}{(x+5)^2}=f_A(x)\). 2. Differentiate \(F_2\): \(F_2'(x)=\frac{2x(x-1)-x^2}{(x-1)^2}=\frac{x^2-2x}{(x-1)^2}=f_B(x)\). 3. Differentiate \(F_3(x)=4x^{-3}\): \(F_3'(x)=-12x^{-4}=f_C(x)\). 4. Differentiate \(F_4(x)=-x^{-4}\): \(F_4'(x)=4x^{-5}=\frac{4}{x^5}=f_D(x)\).

(A) matches (1) (B) matches (2) (C) matches (3) (D) matches (4)

- Rewrite radicals and quotients using rational exponents. - Increase the exponent by \(1\), then divide by the new exponent. - Differentiate each result to check it.

1. Rewrite \(f^{\prime}(x)=\frac32x^{1/2}\). By the power rule, \(f(x)=x^{3/2}\). 2. Rewrite \(g^{\prime}(x)=\frac23x^{-1/3}\). By the power rule, \(g(x)=x^{2/3}\). 3. Rewrite \(h^{\prime}(x)=x^{-3/2}\). By the power rule, \(h(x)=-2x^{-1/2}=-\frac{2}{\sqrt{x}}\).

a) \(f(x)=x^{3/2}\) b) \(g(x)=x^{2/3}\) c) \(h(x)=-\frac{2}{\sqrt{x}}\)

- Rewrite radical denominators using negative exponents. - Integrate each term separately. - A constant derivative of \(-1\) comes from the term \(-x\).

1. By the power rule, \(f(x)=\frac{4/5}{2/5}x^{2/5}=2x^{2/5}\). 2. Rewrite the first term as \(\frac34x^{-3/4}\). Integrating term by term gives \(g(x)=3x^{1/4}-x\).

a) \(f(x)=2x^{2/5}\) b) \(g(x)=3\sqrt[4]{x}-x\)

- Recall the antiderivative of \(x^{-1}\). - Account for the sign of the logarithm's argument when \(x<0\). - Substitute each given function value into the general antiderivative to find \(C\). - Use basic logarithm values when simplifying.

1. On \(x<0\), the family of antiderivatives is \(F(x)=\frac{2}{3}\ln(-x)-4x+C\). 2. Use \(F_1(-1)=0\): \(\frac{2}{3}\ln(1)+4+C=0\). Since \(\ln(1)=0\), \(C=-4\). Thus, \(F_1(x)=\frac{2}{3}\ln(-x)-4x-4\). 3. Use \(F_2(-e)=1\): \(\frac{2}{3}\ln(e)+4e+C=1\). Therefore, \(\frac{2}{3}+4e+C=1\), so \(C=\frac{1}{3}-4e\). 4. Hence, \(F_2(x)=\frac{2}{3}\ln(-x)-4x+\frac{1}{3}-4e\).

1. \(F(x)=\frac{2}{3}\ln(-x)-4x+C\) 2. \(F_1(x)=\frac{2}{3}\ln(-x)-4x-4\) 3. \(F_2(x)=\frac{2}{3}\ln(-x)-4x+\frac{1}{3}-4e\)

- Differentiate \(G\) and compare it with \(g\). - Use the chain rule for \(\ln(-x)\). - For \(x>0\), integrate the reciprocal and polynomial terms. - Use the given function value to determine \(C\).

1. For \(x<0\), \(G'(x)=5\left(\frac{1}{-x}\right)(-1)+\frac{1}{2}x=\frac{5}{x}+\frac{1}{2}x=g(x)\). 2. For \(x>0\), the general antiderivative is \(H(x)=5\ln(x)+\frac{1}{4}x^2+C\). 3. Apply \(H(1)=0.25\): \(5\ln(1)+\frac{1}{4}+C=0.25\). 4. Since \(\ln(1)=0\) and \(\frac{1}{4}=0.25\), \(C=0\). Therefore, \(H(x)=5\ln(x)+\frac{1}{4}x^2\).

1. \(G'(x)=g(x)\) for \(x<0\). 2. \(H(x)=5\ln(x)+\frac{1}{4}x^2\)

- Check the derivative separately on each interval of the domain. - Use the chain rule for the logarithm in the region \(x<-4\). - Does a disconnected domain require the same additive constant on both components? - Compare the derivative with \(f\) on each interval.

1. For \(x>-4\), \(F'(x)=\frac{d}{dx}[\ln(x+4)-2]=\frac{1}{x+4}\). 2. For \(x<-4\), \(F'(x)=\frac{d}{dx}[\ln(-x-4)+5]=\frac{1}{-x-4}(-1)=\frac{1}{x+4}\). 3. Thus, \(F'(x)=f(x)\) on both components of the domain. 4. The different additive constants are allowed because the domain consists of two disconnected intervals. Therefore, \(F\) is an antiderivative of \(f\).

Yes. \(F^{\prime}(x)=f(x)\) on both \((-\infty, -4)\) and \((-4, \infty)\).

- Use the product rule. - Include the chain-rule factor when differentiating \(e^{2x}\). - Differentiate the candidates and compare each result with \(f\).

1. Differentiate \(F_1(x)=xe^{2x}+7\) using the product and chain rules. 2. \(F_1'(x)=e^{2x}+2xe^{2x}=(2x+1)e^{2x}=f(x)\). 3. The other derivatives are \(F_2'(x)=(2x+3)e^{2x}\), \(F_3'(x)=2e^{2x}\), and \(F_4'(x)=(2x+2)e^x\). 4. Therefore, only \(F_1\) is an antiderivative of \(f\).

\(F_1(x)=xe^{2x}+7\)

- Differentiate \(F\) to verify the antiderivative relationship. - Use the chain rule for a squared logarithm. - How do all antiderivatives of the same function differ? - Use the logarithm value at the given input when solving for \(C\).

1. Differentiate using the chain rule: \(F'(x)=\frac{1}{2}\cdot2\ln(x)\cdot\frac{1}{x}=\frac{\ln(x)}{x}=f(x)\). 2. Every antiderivative has the form \(G(x)=\frac{1}{2}[\ln(x)]^2+C\). 3. Use \(G(e)=0\): \(\frac{1}{2}[\ln(e)]^2+C=0\). 4. Since \(\ln(e)=1\), \(C=-\frac{1}{2}\). Thus, \(G(x)=\frac{1}{2}[\ln(x)]^2-\frac{1}{2}\).

1. \(F'(x)=\frac{\ln(x)}{x}=f(x)\). 2. \(G(x)=\frac{1}{2}[\ln(x)]^2-\frac{1}{2}\)

- Differentiate the proposed antiderivative. - Use the product rule on the term containing \(x^3\ln(x)\). - All antiderivatives differ by a constant. - Substitute the point to determine that constant.

1. Rewrite \(F(x)=\frac{1}{3}x^3\ln(x)-\frac{1}{9}x^3\). 2. Differentiate using the product rule: \(F'(x)=x^2\ln(x)+\frac{1}{3}x^2-\frac{1}{3}x^2=x^2\ln(x)=f(x)\). 3. Every antiderivative has the form \(H(x)=\frac{1}{9}x^3(3\ln(x)-1)+C\). 4. Use \(H(1)=1\): \(-\frac{1}{9}+C=1\), so \(C=\frac{10}{9}\). 5. Therefore, \(H(x)=\frac{1}{9}x^3(3\ln(x)-1)+\frac{10}{9}\).

1. \(F'(x)=x^2\ln(x)=f(x)\). 2. \(H(x)=\frac{1}{9}x^3(3\ln(x)-1)+\frac{10}{9}\)

- Pay attention to the logarithm's argument because \(x<0\). - Use the chain rule to check the derivative of \(\ln(-x)\). - Different constants give different antiderivatives.

1. Integrate term by term. An antiderivative of \(x\) is \(\frac12x^2\). 2. On \(x<0\), an antiderivative of \(\frac1x\) is \(\ln(-x)\). Therefore, a general antiderivative is \(F(x)=\frac12x^2-\frac14\ln(-x)+C\). 3. Choosing \(C=0\) and \(C=-1\) gives two different antiderivatives.

\(F_1(x)=\frac12x^2-\frac14\ln(-x)\) and \(F_2(x)=\frac12x^2-\frac14\ln(-x)-1\)

- Recall the derivatives of sine and cosine. - Rewrite expressions with \(x\) in the denominator using negative exponents. - Integrate sums term by term. - Check the sign when working with sine and cosine.

1. Since the derivative of \(\sin(x)\) is \(\cos(x)\), one choice for a) is \(f(x)=5\sin(x)\). 2. Since the derivative of \(\cos(x)\) is \(-\sin(x)\), one choice for b) is \(f(x)=\cos(x)\). 3. Rewrite \(\frac{2}{x^2}\) as \(2x^{-2}\). Reversing the power rule gives \(f(x)=-2x^{-1}=-\frac2x\). 4. Integrating term by term gives \(f(x)=\frac13x^3-\frac12x^2\).

a) \(f(x)=5\sin(x)\) b) \(f(x)=\cos(x)\) c) \(f(x)=-\frac2x\) d) \(f(x)=\frac13x^3-\frac12x^2\)

- Integrate the two terms separately. - Which trigonometric function has derivative \(\sin(x)\)? Pay attention to the sign. - Reverse the power rule for \(4x^7\). - Differentiate your result to verify it.

1. Since the derivative of \(-8\cos(x)\) is \(8\sin(x)\), the first term has antiderivative \(-8\cos(x)\). 2. Reversing the power rule gives an antiderivative of \(4x^7\) as \(\frac12x^8\). 3. Therefore, one possible function is \(g(x)=-8\cos(x)+\frac12x^8\). 4. Differentiating gives \(g^{\prime}(x)=8\sin(x)+4x^7\), as required.

\(g(x)=-8\cos(x)+\frac12x^8\). More generally, \(g(x)=-8\cos(x)+\frac12x^8+C\), where \(C\in\mathbb{R}\).

- Rewrite the fraction using a negative exponent. - Recall which trigonometric function has derivative \(\cos(x)\). - Reverse the power rule, then differentiate your result to check it.

1. Rewrite \(\frac{4}{x^5}\) as \(4x^{-5}\). 2. An antiderivative of \(\cos(x)\) is \(\sin(x)\). 3. Reversing the power rule gives an antiderivative of \(4x^{-5}\) as \(-x^{-4}=-\frac1{x^4}\). 4. Therefore, one possible function is \(g(x)=\sin(x)-\frac1{x^4}\).

\(g(x)=\sin(x)-\frac1{x^4}\) for \(x\ne0\)

- Differentiate an antiderivative to recover the original function. - How do antiderivatives of the same function differ? - What happens when you differentiate their difference?

1. Since \(F\) is an antiderivative of \(f\), \(f(x)=F'(x)=2x^3-2x\). 2. Every antiderivative of \(f\) has the form \(G(x)=\frac{1}{2}x^4-x^2+C\). 3. Use \(G(2)=10\): \(8-4+C=10\), so \(C=6\). Thus, \(G(x)=\frac{1}{2}x^4-x^2+6\). 4. If \(F\) and \(G\) are both antiderivatives of \(f\), then \((G-F)'=G'-F'=f-f=0\). A function with derivative \(0\) on an interval is constant, so \(G-F=C\).

a) \(f(x)=2x^3-2x\) b) \(G(x)=\frac{1}{2}x^4-x^2+6\) c) The derivative of \(G-F\) is \(0\), so \(G-F\) is constant.

- Rewrite terms with \(x\) in the denominator using negative exponents. - Split a quotient with a sum in the numerator into separate terms. - Apply the reverse power rule term by term.

1. For a), rewrite the function as \(x^4-2x^{-3}+1\). Integrating term by term gives \(F(x)=\frac15x^5+x^{-2}+x=\frac15x^5+\frac1{x^2}+x\). 2. For b), simplify first: \(\frac{x^3+5}{x^2}=x+5x^{-2}\). Integrating term by term gives \(F(x)=\frac12x^2-5x^{-1}=\frac12x^2-\frac5x\).

a) \(F(x)=\frac15x^5+\frac1{x^2}+x\) b) \(F(x)=\frac12x^2-\frac5x\)

- Rewrite radicals using rational exponents. - Simplify the quotient in b) before integrating. - Apply the reverse power rule carefully to fractional exponents.

1. For a), rewrite the function as \(3x^{1/2}-x^{-1/2}\). Integrating term by term gives \(F(x)=2x^{3/2}-2x^{1/2}=2x^{3/2}-2\sqrt{x}\). 2. For b), simplify first: \(\frac{\sqrt{x}+2}{\sqrt{x}}=1+2x^{-1/2}\). Integrating gives \(F(x)=x+4x^{1/2}=x+4\sqrt{x}\).

a) \(F(x)=2x^{3/2}-2\sqrt{x}\) b) \(F(x)=x+4\sqrt{x}\)

- First write the general antiderivative with a constant \(C\). - Substitute the given input and output values. - Solve each resulting equation for its constant.

1. The general antiderivative is \(F(x)=x^3+2x^2+C\). 2. From \(F_1(0)=0\), \(C_1=0\), so \(F_1(x)=x^3+2x^2\). 3. From \(F_2(0)=10\), \(C_2=10\), so \(F_2(x)=x^3+2x^2+10\). 4. From \(F_3(1)=5\), \(1+2+C_3=5\), so \(C_3=2\) and \(F_3(x)=x^3+2x^2+2\).

1. \(F_1(x)=x^3+2x^2\) 2. \(F_2(x)=x^3+2x^2+10\) 3. \(F_3(x)=x^3+2x^2+2\)

- Divide each term in the numerator by \(x^2\) before integrating. - Rewrite reciprocal powers using negative exponents. - Different antiderivatives differ by a constant.

1. Simplify the function: \(f(x)=6x^2-3x^{-2}+5x^4\). 2. Integrating term by term gives the general antiderivative \(F(x)=2x^3+3x^{-1}+x^5+C=x^5+2x^3+\frac3x+C\). 3. Choose two different constants, such as \(C=0\) and \(C=5\).

\(F_1(x)=x^5+2x^3+\frac3x\) \(F_2(x)=x^5+2x^3+\frac3x+5\)

- Expand the squared binomial first. - Rewrite the radical in the denominator using a negative exponent. - Integrate the resulting sum term by term. - Different constants produce different antiderivatives.

1. Expand and rewrite the function: \(g(x)=4x^4+4x^2+1-4x^{-1/2}\). 2. Integrating term by term gives \(G(x)=\frac45x^5+\frac43x^3+x-8\sqrt{x}+C\). 3. Choosing \(C=0\) and \(C=1\) gives two different antiderivatives.

\(G_1(x)=\frac45x^5+\frac43x^3+x-8\sqrt{x}\) \(G_2(x)=\frac45x^5+\frac43x^3+x-8\sqrt{x}+1\)

- Treat the parameter \(k\) as a constant. - Apply the reverse power rule to each term. - Differentiate your result to verify it.

1. Treat \(k\) as a constant and integrate each term with respect to \(x\). 2. The first term gives \(\frac{1}{k+1}\cdot\frac{x^{k+1}}{k+1}=\frac{x^{k+1}}{(k+1)^2}\). 3. The second term gives \(-k\cdot\frac{x^k}{k}=-x^k\). 4. Therefore, \(H(x)=\frac{x^{k+1}}{(k+1)^2}-x^k\).

\(H(x)=\frac{x^{k+1}}{(k+1)^2}-x^k\)

- Rewrite each function as a sum of powers before integrating. - Expand products and squared binomials when needed. - Rewrite terms with \(x\) in the denominator using negative exponents. - Treat \(p\) as a constant when integrating with respect to \(x\).

1. For a), expand to \(9x^2-6x+1\), then integrate term by term: \(F(x)=3x^3-3x^2+x\). 2. For b), expand to \(\frac12x^3+2x^2\), then integrate: \(F(x)=\frac18x^4+\frac23x^3\). 3. For c), simplify to \(x^2-2+6x^{-2}\), then integrate: \(F(x)=\frac13x^3-2x-\frac6x\). 4. For d), treat \(p\) as a constant. Expanding gives \(px^2-2p^2x+p^3\), so \(F(x)=\frac p3x^3-p^2x^2+p^3x\).

a) \(F(x)=3x^3-3x^2+x\) b) \(F(x)=\frac18x^4+\frac23x^3\) c) \(F(x)=\frac13x^3-2x-\frac6x\) d) \(F(x)=\frac p3x^3-p^2x^2+p^3x\)

- Expand powers of binomials before integrating. - Use exponent rules to simplify quotients. - Rewrite radicals and reciprocal powers using rational or negative exponents. - Differentiate each result to check it.

1. For a), expand to \(1-3x+3x^2-x^3\). Integrating term by term gives \(F(x)=x-\frac32x^2+x^3-\frac14x^4\). 2. For b), simplify to \(4x^3-2x^{-1/2}\). Integrating gives \(F(x)=x^4-4\sqrt{x}\). 3. For c), expand to \(x^4+4x^2+4\). Integrating gives \(F(x)=\frac15x^5+\frac43x^3+4x\). 4. For d), simplify to \(x-3x^{-2}+2x^{-4}\). Integrating gives \(F(x)=\frac12x^2+\frac3x-\frac{2}{3x^3}\).

a) \(F(x)=x-\frac32x^2+x^3-\frac14x^4\) b) \(F(x)=x^4-4\sqrt{x}\) c) \(F(x)=\frac15x^5+\frac43x^3+4x\) d) \(F(x)=\frac12x^2+\frac3x-\frac{2}{3x^3}\)

- Recall the antiderivatives of sine and cosine. - Rewrite reciprocal powers and radicals using exponents. - Integrate each term separately and keep constant factors.

1. For a), integrate term by term: \(F(x)=-5\cos(x)+\frac15x^5\). 2. For b), rewrite \(\frac3{x^2}\) as \(3x^{-2}\). Integrating gives \(F(x)=-\frac3x-2\sin(x)\). 3. For c), integrate inside the parentheses and retain the factor \(4\): \(F(x)=4\sin(x)+8\sqrt{x}\).

a) \(F(x)=-5\cos(x)+\frac15x^5\) b) \(F(x)=-\frac3x-2\sin(x)\) c) \(F(x)=4\sin(x)+8\sqrt{x}\)

- First find the general antiderivative with a constant \(C\). - Substitute the coordinates of the given point. - Use \(\cos(0)=1\) to solve for \(C\).

1. The general antiderivative is \(F(x)=-\cos(x)-x^4+C\). 2. Use the condition \(F(0)=2\): \(-\cos(0)+C=2\). 3. Since \(\cos(0)=1\), \(-1+C=2\), so \(C=3\). 4. Therefore, \(F(x)=-\cos(x)-x^4+3\).

\(F(x)=-\cos(x)-x^4+3\)

- Account for the chain-rule factor when integrating \(e^{2x}\). - Evaluate the exponential term at the given input. - Use the given function value to find \(C\). - Differentiate your result to check it.

1. Integrate term by term: \(F(x)=2e^{2x}-2x^3+C\). 2. Apply \(F(0)=5\): \(2e^0+C=5\). 3. Thus, \(2+C=5\), so \(C=3\). 4. Therefore, \(F(x)=2e^{2x}-2x^3+3\).

\(F(x)=2e^{2x}-2x^3+3\)

- Integrate each polynomial term using the power rule. - What equation follows from a zero at \(x=2\)? - Include the integration constant in the general antiderivative. - Solve the resulting equation for \(C\).

1. The general antiderivative is \(F(x)=\frac{1}{8}x^4-x^3+5x+C\). 2. The zero at \(x=2\) means \(F(2)=0\): \(2-8+10+C=0\). 3. Therefore, \(4+C=0\), so \(C=-4\). 4. Thus, \(F(x)=\frac{1}{8}x^4-x^3+5x-4\).

\(F(x)=\frac{1}{8}x^4-x^3+5x-4\)

- Rewrite the reciprocal term using a negative exponent. - Apply the power rule for antiderivatives carefully. - Use the condition \(F(2)=0\). - Solve for the remaining constant.

1. Rewrite \(f(x)=4x^{-3}+\frac{1}{2}\). 2. Integrate: \(F(x)=-2x^{-2}+\frac{1}{2}x+C=-\frac{2}{x^2}+\frac{1}{2}x+C\). 3. Use \(F(2)=0\): \(-\frac{2}{4}+1+C=0\). 4. Thus, \(\frac{1}{2}+C=0\), so \(C=-\frac{1}{2}\). 5. Therefore, \(F(x)=-\frac{2}{x^2}+\frac{1}{2}x-\frac{1}{2}\).

\(F(x)=-\frac{2}{x^2}+\frac{1}{2}x-\frac{1}{2}\)

- How do antiderivatives of the same function differ? - Differentiate \(F\) to recover \(f\). - Use the point on \(G\) to determine the constant.

1. Recover \(f\) by differentiating \(F\): \(f(x)=F'(x)=-\frac{1}{x^2}+2x\). 2. Since \(F\) and \(G\) are antiderivatives of the same function, \(G(x)=F(x)+C=\frac{1}{x}+x^2+C\). 3. Use \(G(1)=5\): \(1+1+C=5\), so \(C=3\). 4. Therefore, \(G(x)=\frac{1}{x}+x^2+3\).

\(f(x)=-\frac{1}{x^2}+2x\) \(G(x)=\frac{1}{x}+x^2+3\)

- Recall the antiderivatives related to square-root and exponential functions. - Substitute the given input into the general antiderivative. - Solve the resulting equation for \(C\).

1. The general antiderivative is \(F(x)=\sqrt{x}+e^x+C\). 2. Apply \(F(1)=e+4\): \(1+e+C=e+4\). 3. Therefore, \(C=3\). 4. Thus, \(F(x)=\sqrt{x}+e^x+3\).

\(F(x)=\sqrt{x}+e^x+3\)

- For \(e^{ax+b}\), divide by the derivative of the exponent. - Integrate sums term by term. - Differentiate each result to check the inner-factor adjustment.

1. For a), account for the inner derivative \(2\): \(F(x)=2e^{2x}\). 2. For b), the inner derivative of \(5-x\) is \(-1\), so an antiderivative of \(e^{5-x}\) is \(-e^{5-x}\). Therefore, \(F(x)=-e^{5-x}+\frac13x^3\). 3. For c), account for the inner derivative \(4\): \(F(x)=\frac18e^{4x-2}-3x\).

a) \(F(x)=2e^{2x}\) b) \(F(x)=-e^{5-x}+\frac13x^3\) c) \(F(x)=\frac18e^{4x-2}-3x\)

- First find the general antiderivative with a constant \(C\). - Account for the inner derivative of \(-x\). - Substitute the coordinates of the given point to determine \(C\).

1. Accounting for the inner derivative \(-1\), the general antiderivative is \(F(x)=-3e^{-x}+2x+C\). 2. Use \(F(0)=1\): \(-3e^0+C=1\). 3. Since \(e^0=1\), \(-3+C=1\), so \(C=4\). 4. Therefore, \(F(x)=-3e^{-x}+2x+4\).

\(F(x)=-3e^{-x}+2x+4\)

- Which logarithmic function differentiates to \(\frac{1}{x}\) when \(x<0\)? - Different antiderivatives differ only by constants. - How is the slope of an antiderivative related to \(f\)? - Evaluate \(f\) at the stated input.

1. On \(x<0\), the family of antiderivatives is \(F(x)=\ln(-x)+C\). 2. Three examples are \(F_1(x)=\ln(-x)\), \(F_2(x)=\ln(-x)+2\), and \(F_3(x)=\ln(-x)-1\). 3. Every antiderivative satisfies \(F'(x)=f(x)\), so the slope at \(x=-2\) is always \(f(-2)\). 4. Since \(f(-2)=-\frac{1}{2}\), all the antiderivatives have slope \(-\frac{1}{2}\) there.

a) \(F_1(x)=\ln(-x)\), \(F_2(x)=\ln(-x)+2\), and \(F_3(x)=\ln(-x)-1\) b) All have slope \(f(-2)=-\frac{1}{2}\).

- Read the slope and y-intercept of the line from the graph. - Integrate the derivative and include a constant \(C\). - Use the given point to determine \(C\).

1. The graph of \(f'\) is a line with y-intercept \(2\) and zero at \(x=4\). Its slope is \(\frac{0-2}{4-0}=-\frac{1}{2}\), so \(f'(x)=-\frac{1}{2}x+2\). 2. Integrate to obtain \(f(x)=-\frac{1}{4}x^2+2x+C\). 3. Use \(f(2)=1\): \(-1+4+C=1\), so \(C=-2\). 4. Therefore, \(f(x)=-\frac{1}{4}x^2+2x-2\).

\(f(x)=-\frac{1}{4}x^2+2x-2\)

- Match local extrema of a graph with zeros of its derivative. - Compare increasing and decreasing intervals with the sign of the derivative. - Use the polynomial degrees to check the order \(F\), \(f\), and \(f'\).

1. Graph B has a local maximum at \(x = 2\) and a local minimum at \(x = 6\). Graph C is zero at those values, is negative on \((2, 6)\), and is positive outside that interval. Therefore, graph C is the derivative of graph B. 2. Graph C has a local minimum at \(x = 4\). Graph A is zero at \(x = 4\), is negative for \(x < 4\), and is positive for \(x > 4\). Therefore, graph A is the derivative of graph C. 3. Since \(F' = f\) and \(f'\) is the derivative of \(f\), graph B represents \(F\), graph C represents \(f\), and graph A represents \(f'\).

A \(\rightarrow f'\) B \(\rightarrow F\) C \(\rightarrow f\)

- Use the product rule to differentiate \(x\ln(x)\). - Use the first and second derivatives to classify the critical point. - Write the tangent line at a general value \(u\), then substitute \(P\). - To verify an antiderivative, differentiate it and compare the result with \(f\).

1. Differentiate using the product rule: \(f'(x)=\ln(x)+1\). Setting \(f'(x)=0\) gives \(\ln(x)=-1\), so \(x=\frac{1}{e}\). Since \(f''(x)=\frac{1}{x}>0\) for \(x>0\), this point is a local minimum. Its y-coordinate is \(f\left(\frac{1}{e}\right)=-\frac{1}{e}\). 2. Let the point of tangency have x-coordinate \(u>0\). The tangent line is \(y=(\ln(u)+1)(x-u)+u\ln(u)=(\ln(u)+1)x-u\). Because it passes through \(P(0,-1)\), \(-1=-u\), so \(u=1\). Therefore, the tangent line is \(y=x-1\). 3. Differentiate \(F\): \(F'(x)=x\ln(x)+\frac{1}{2}x-\frac{1}{2}x=x\ln(x)=f(x)\). Therefore, \(F\) is an antiderivative of \(f\).

a) \(\left(\frac{1}{e}, -\frac{1}{e}\right)\) b) \(y=x-1\) c) \(F'(x)=x\ln(x)=f(x)\), so \(F\) is an antiderivative of \(f\).

- Use the intercepts and vertex of the parabola to determine \(f^{\prime}\). - Integrate \(f^{\prime}\) and include a constant of integration. - Use the point on \(f\) to determine the constant.

1. The derivative has zeros at \(x=-2\) and \(x=2\) and a y-intercept of \(4\). Therefore, \(f^{\prime}(x)=-(x-2)(x+2)=4-x^2\). 2. Integrating gives \(f(x)=4x-\frac13x^3+C\). 3. Use \(f(3)=2\): \(2=12-9+C\), so \(C=-1\). 4. Thus, \(f(x)=-\frac13x^3+4x-1\).

\(f(x)=-\frac13x^3+4x-1\)

- The sign of \(f\) determines whether \(F\) increases or decreases. - The increasing or decreasing behavior of \(f\) determines the concavity of \(F\). - Use the initial condition to determine the constant of integration.

1. Since \(f(x)=\frac{2}{x+1}>0\) for \(x>-1\), every antiderivative is strictly increasing on that interval. 2. Integrating gives \(F(x)=2\ln(x+1)+C\). The condition \(F(0)=0\) gives \(C=0\), so \(F(x)=2\ln(x+1)\). 3. Because \(f\) is decreasing, the slope of \(F\) decreases, so \(F\) is concave down on \((-1,\infty)\). 4. As \(x\to-1^+\), \(F(x)\to-\infty\). As \(x\to\infty\), \(F\) continues to increase but becomes progressively flatter because \(f(x)\to0\).

\(F(x)=2\ln(x+1)\). Its graph passes through \((0,0)\), is strictly increasing and concave down on \((-1,\infty)\), approaches \(-\infty\) as \(x\to-1^+\), and becomes flatter as \(x\to\infty\).

- Integrate \(x^{1/2}\) using the power rule. - Use \(F(0)=-1\) to determine the constant. - Use the sign and monotonicity of \(f\) to describe the shape of \(F\).

1. Integrating gives \(F(x)=\frac23x^{3/2}+C\). 2. The condition \(F(0)=-1\) gives \(C=-1\), so \(F(x)=\frac23x^{3/2}-1\). 3. Since \(f(x)>0\) for \(x>0\), \(F\) is strictly increasing there. Since \(f\) is increasing, the slope of \(F\) increases, so \(F\) is concave up. 4. The graph begins at \((0,-1)\) with a horizontal tangent. Useful points include \((1,-\frac13)\) and \((4,\frac{13}{3})\).

\(F(x)=\frac23x^{3/2}-1\). The graph starts at \((0,-1)\) with a horizontal tangent and is increasing and concave up for \(x>0\).

- How are \(f\) and an antiderivative \(F\) related? - Where can \(F\) have local extrema? - How can the derivative of \(f\) distinguish minima from maxima of \(F\)? - How does the stated minimum value determine \(C\)?

1. The general antiderivative is \(F(x)=\frac{1}{4}x^4-2x^2+C\). 2. Critical points of \(F\) occur where \(F'(x)=f(x)=0\): \(x(x^2-4)=0\), so \(x=-2,0,2\). 3. Since \(F''(x)=f'(x)=3x^2-4\), the points at \(x=\pm2\) are local minima and the point at \(x=0\) is a local maximum. 4. Use the required minimum value: \(F(2)=5\). Then \(\frac{1}{4}(2)^4-2(2)^2+C=5\), so \(4-8+C=5\) and \(C=9\). 5. Therefore, \(F(x)=\frac{1}{4}x^4-2x^2+9\).

\(F(x)=\frac{1}{4}x^4-2x^2+9\)

- Which function has derivative \(\frac{1}{x}\)? - Set \(F'(x)=f(x)\) equal to zero to find critical points of \(F\). - Use logarithm properties when simplifying the condition. - The constant \(C\) shifts the graph vertically.

1. For \(x>0\), the general antiderivative is \(F(x)=\frac{1}{4}x^2-2\ln(x)+C\). 2. Critical points of \(F\) satisfy \(F'(x)=f(x)=0\): \(\frac{1}{2}x-\frac{2}{x}=0\), so \(x^2=4\). Because \(x>0\), \(x=2\). 3. Since \(F''(x)=f'(x)=\frac{1}{2}+\frac{2}{x^2}>0\), \(F\) has a local minimum at \(x=2\). 4. Apply the minimum value: \(F(2)=5-\ln(4)\). This gives \(1-2\ln(2)+C=5-\ln(4)\). 5. Because \(2\ln(2)=\ln(4)\), \(C=4\). Thus, \(F(x)=\frac{1}{4}x^2-2\ln(x)+4\).

\(F(x)=\frac{1}{4}x^2-2\ln(x)+4\)

- Which parts of the two terms are derivatives of each other? - Recall the relationship between \(\sqrt{x}\) and \(\frac{1}{2\sqrt{x}}\). - How does the negative sign relate to the derivative of cosine? - Try differentiating a product involving \(\sqrt{x}\) and \(\cos(x)\).

1. Look for a product whose derivative has the two given terms. 2. Let \(u(x)=\sqrt{x}\) and \(v(x)=\cos(x)\). Then \(u'(x)=\frac{1}{2\sqrt{x}}\) and \(v'(x)=-\sin(x)\). 3. By the product rule, \(\frac{d}{dx}[\sqrt{x}\cos(x)]=\frac{\cos(x)}{2\sqrt{x}}-\sqrt{x}\sin(x)=f(x)\). 4. Thus, one antiderivative is \(F(x)=\sqrt{x}\cos(x)\).

\(F(x)=\sqrt{x}\cos(x)\)

- Differentiate \(F\) and compare the result with \(f\). - The derivative of \(F\) determines the extrema of \(F\). - Use the second derivative of each function to analyze inflection points. - On negative inputs, pay attention to the signs of odd powers.

1. Differentiate \(F\): \(F'(x)=0.5-\frac{2}{x^2}=\frac{x^2-4}{2x^2}=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 2. Local extrema of \(F\) occur where \(F'(x)=f(x)=0\). Thus \(x^2-4=0\), so \(x=-2\) and \(x=2\). Since \(F''(x)=f'(x)=\frac{4}{x^3}\), \(F''(-2)=-\frac{1}{2}<0\) and \(F''(2)=\frac{1}{2}>0\). Therefore, \((-2,2)\) is a local maximum and \((2,6)\) is a local minimum. 3. For \(F\), \(F''(x)=\frac{4}{x^3}\) is negative on \((-\infty,0)\) and positive on \((0,\infty)\). Although the concavity differs on the two domain intervals, \(x=0\) is not in the domain, so \(F\) has no inflection point. For \(f\), \(f''(x)=-\frac{12}{x^4}<0\) everywhere on its domain, so its graph is concave down on both domain intervals and has no inflection point. 4. On \((-\infty,0)\), \(f'(x)=\frac{4}{x^3}<0\), so \(f\) is strictly decreasing. Also, \(f''(x)=-\frac{12}{x^4}<0\), so \(f'\) is strictly decreasing. Finally, \(f'''(x)=\frac{48}{x^5}<0\), so \(f''\) is strictly decreasing. All three functions are strictly decreasing on that interval.

a) \(F'(x)=f(x)\), so \(F\) is an antiderivative of \(f\). b) Local maximum: \((-2,2)\). Local minimum: \((2,6)\). c) Neither graph has an inflection point. d) \(f\), \(f'\), and \(f''\) are all strictly decreasing on \((-\infty,0)\).

- Rewrite the rational expression before differentiating. - Separate the linear part from the term that approaches zero to find the slant asymptote. - Use the sign of each second derivative to determine concavity. - Use the sign of each first derivative to determine monotonicity.

1. Rewrite \(g(x)=2x-\frac{1}{x}\). Then \(g'(x)=2+\frac{1}{x^2}=h(x)\), so \(g\) is an antiderivative of \(h\). 2. The function is undefined at \(x=0\), and its values become unbounded there, so \(x=0\) is a vertical asymptote. Since \(g(x)=2x-\frac{1}{x}\) and \(-\frac{1}{x}\to0\) as \(|x|\to\infty\), the slant asymptote is \(y=2x\). 3. For \(g\), \(g''(x)=h'(x)=-\frac{2}{x^3}\). This is positive when \(x<0\) and negative when \(x>0\), so \(g\) is concave up on \((-\infty,0)\) and concave down on \((0,\infty)\). For \(h\), \(h''(x)=\frac{6}{x^4}>0\), so \(h\) is concave up on both \((-\infty,0)\) and \((0,\infty)\). 4. On \((-\infty,0)\), \(g'(x)=2+\frac{1}{x^2}>0\), so \(g\) is strictly increasing. Also, \(h'(x)=-\frac{2}{x^3}>0\), so \(h\) is strictly increasing. Finally, \(h''(x)=\frac{6}{x^4}>0\), so \(h'\) is strictly increasing. All three are strictly increasing on that interval.

a) \(g'(x)=2+\frac{1}{x^2}=h(x)\), so \(g\) is an antiderivative of \(h\). b) Vertical asymptote: \(x=0\). Slant asymptote: \(y=2x\). c) \(g\) is concave up on \((-\infty,0)\) and concave down on \((0,\infty)\). The graph of \(h\) is concave up on \((-\infty,0)\cup(0,\infty)\). d) \(g\), \(h\), and \(h'\) are all strictly increasing on \((-\infty,0)\).

- Set the logarithm input greater than \(0\) to find the domain. - Use \(x=0\) for the \(y\)-intercept and set \(g(x)=0\) for the \(x\)-intercept. - Apply the graph transformations one at a time. - Use the product rule and chain rule to differentiate \(G\). - Add a constant and use the point \((4, 0)\).

1. The logarithm requires \(5-x>0\), so the domain is \((-\infty, 5)\). As \(x\to5^-\), \(\ln(5-x)\to-\infty\), so \(g(x)\to\infty\). As \(x\to-\infty\), \(\ln(5-x)\to\infty\), so \(g(x)\to-\infty\). 2. For the \(y\)-intercept, \(g(0)=2-\ln(5)\), giving \((0, 2-\ln(5))\). For the \(x\)-intercept, solve \(2-\ln(5-x)=0\). Then \(5-x=e^2\), so the intercept is \((5-e^2, 0)\). 3. Starting with \(y=\ln(x)\), reflect across the \(y\)-axis, shift right \(5\) units, reflect across the \(x\)-axis, and shift up \(2\) units. The first reflection changes an increasing graph to a decreasing graph, and the second reflection changes it back to an increasing graph. Translations do not change monotonicity, so \(g\) is strictly increasing. 4. Differentiate: \(G'(x)=3-\ln(5-x)-1=2-\ln(5-x)=g(x)\). Thus, the general antiderivative is \(G(x)+C\). 5. Requiring the graph to pass through \((4, 0)\) gives \(0=3(4)+(5-4)\ln(1)+C=12+C\), so \(C=-12\). The required antiderivative is \(3x+(5-x)\ln(5-x)-12\).

a) Domain: \((-\infty, 5)\); \(\lim_{x\to5^-}g(x)=\infty\); \(\lim_{x\to-\infty}g(x)=-\infty\) b) \(x\)-intercept: \((5-e^2, 0)\); \(y\)-intercept: \((0, 2-\ln(5))\) c) Reflect across the \(y\)-axis, shift right \(5\), reflect across the \(x\)-axis, and shift up \(2\). The resulting function is strictly increasing. d) \(3x+(5-x)\ln(5-x)-12\)

- Distinguish between a zero where the graph crosses the x-axis and a zero where it only touches. - Use the y-intercept to determine \(a\). - Integrate term by term and use the given point to find the constant. - A zero of \(f\) without a sign change gives \(F\) a horizontal tangent but not an extremum.

1. The graph touches the x-axis at \(x=2\), so \(x_1=2\) is a double zero. It crosses at \(x=-1\), so \(x_2=-1\). Since \(f(0)=-2\), \(-2=a(0-2)^2(0+1)=4a\), giving \(a=-\frac12\). Thus, \(f(x)=-\frac12(x-2)^2(x+1)=-\frac12x^3+\frac32x^2-2\). 2. Integrating gives \(F(x)=-\frac18x^4+\frac12x^3-2x+C\). The condition \(F(2)=0\) gives \(-2+4-4+C=0\), so \(C=2\). Therefore, \(F(x)=-\frac18x^4+\frac12x^3-2x+2\). 3. Since \(f(2)=F^{\prime}(2)=0\), the graph of \(F\) has a horizontal tangent at \(x=2\). The function \(f\) does not change sign there, so \(F\) does not have a local extremum. Because \(f\) has a local maximum at \(x=2\), \(F\) changes concavity there. Thus, \(F\) has a stationary inflection point at \(x=2\).

a) \(a=-\frac12\), \(x_1=2\), \(x_2=-1\); \(f(x)=-\frac12x^3+\frac32x^2-2\) b) \(F(x)=-\frac18x^4+\frac12x^3-2x+2\) c) A stationary inflection point at \(x=2\)

- Calculate the y-intercept, zeros, and horizontal asymptote of \(F\). - Inflection points of \(F\) correspond to extrema of \(F^{\prime}=f\). - A zero of \(f\) occurs where \(F\) has a horizontal tangent. - Use the increasing and decreasing behavior of \(F\) to determine the sign of \(f\).

1. The function satisfies \(F(0)=-2\), has zeros at \(x=\pm2\), and approaches \(1\) as \(x\to\pm\infty\). Only Graph 1 has all three features. 2. Since \(f=F^{\prime}\), local extrema of \(f\) occur where \(F^{\prime\prime}=0\), which are inflection points of \(F\). Graph 1 has two inflection points, so \(f\) has two local extrema. 3. Zeros of \(f\) occur where \(F\) has horizontal tangents. Graph 1 has one such point, at \(x=0\). Therefore, \(f\) has the single zero \(x=0\). 4. Differentiation gives \(f(x)=F^{\prime}(x)=\frac{12x}{(x^2+2)^2}\). As \(x\to\pm\infty\), \(f(x)\to0\), so the horizontal asymptote is \(y=0\). 5. The denominator is always positive, so \(f(x)<0\) for \(x<0\) and \(f(x)>0\) for \(x>0\).

a) Graph 1 b) \(f\) has two local extrema. c) \(x=0\) d) The horizontal asymptote is \(y=0\). The graph is below the x-axis for \(x<0\) and above it for \(x>0\).