We will first find all critical numbers. Then we will evaluate the function at each critical number and identify the minimum. There are 4 categories of critical numbers

• Category 1: Endpoints. There are two critical numbers in this category. They are \(x=0\) and \(x=2\);
• Category 2: Values at which the function is not defined. There are no critical values in this category since the function is continuous on the entire interval. The denominator \(x^2+1\) is never \(0\).
• Category 3: Values at which the function is not differentiable. There are no critical values in this category. The function is differentiable and its derivative is \begin{eqnarray*}f'(x)&=&\frac{(2x)'\cdot(x^2+1)-2x\cdot\left(x^2+1\right)'}{\left(x^2+1\right)^2} \\&=&\frac{2x^2+2-2x\cdot 2x}{\left(x^2+1\right)^2}\\ &=&2\cdot\frac{1-x^2}{\left(x^2+1\right)^2}. \end{eqnarray*}
• Category 4: Values at which the derivative is equal to \(0\). We need to solve the equation \(f'(x)=0\). The equation becomes \[2\cdot\frac{1-x^2}{\left(x^2+1\right)^2}=0.\] After multiplying both sides by \(\left(x^2+1\right)^2\) and canceling by \(2\) we obtain \(1-x^2=0\). This equation is equivalent to \((1-x)(1+x)=0\). The solutions are \(x=1\) and \(x=-1\). The value \(x=-1\) is not in the domain of the function. Therefore the only critical value in the category \(4\) is \(x=1\).

Therefore, there are \(3\) critical values of our function. They are \(x=0\), \(x=1\), and \(x=2\). The values of the function are \(f(0)=0\), \(f(1)=1\), and \(f(2)=\frac45\). The minimum is \(0\). Therefore, the correct answer is B.

We will use the equation \[\left(f^{-1}\right)'(z)=\frac1{f'\left(f^{-1}(z)\right)}.\] Substituting \(z=5\) we obtain \begin{eqnarray*} \left(f^{-1}\right)'(5)&=&\frac1{f'\left(f^{-1}(5)\right)}=\frac1{f'(-1)}=\frac14. \end{eqnarray*} We have used that \(f^{-1}(5)=-1\) because from the formulation of the problem we know that \(f(-1)=5\). The correct answer is D.

The average cost is given by \[f(x)=\frac{C(x)}{x}=2x^2-20x+100.\] The only critical value of the function \(f(x)\) is the solution of the equation \(f'(x)=0\). The equation is equivalent to \(4x-20=0\). Therefore, the critical value is \(x=5\). The function goes to infinity as \(x\to\pm\infty\).Therefore, \(f(5)\) is the minimum. The correct answer is A.

We will take logarithms of both sides of the equation. The equation becomes \[\ln\left(\left(\frac13\right)^{1-x}\right)=\ln9.\] We will now use that \(\ln A^B=B\ln A\) and that \(9=3^2\). The equation is equivalent to \[(1-x)\ln\frac13=2\ln 3.\] Since \(\ln\frac{C}{D}=\ln C-\ln D\) we have that \(\ln\frac13=\ln1-\ln3=0-\ln 3\). Therefore, the original equation is equivalent to \(-(1-x)\ln3 =2 \ln3\). Dividing both sides by \(\ln3\) gives us \(x-1=2\), which implies that \(x=3\). Thus, the correct answer is E.

Observe first that we must have \(x> 0\) and \(x-7> 0\). Otherwise, the logarithms would not be defined. We will use that \(\log_AB+\log_AC=\log_A(BC)\) to transform the original equation to \[\log_2(x(x-7))=3.\] The last equation is equivalent to \[2^{\log_2\left(x(x-7)\right)}=2^3.\] Since \(2^{\log_2Q}=Q\), we obtain \[x(x-7)=8,\] which is equivalent to \(x^2-7x-8=0\). The last quadratic equation is the same as \((x-8)(x+1)=0\). The solutions are \(x=8\) and \(x=-1\). Since we must have \(x> 0\) and \(x-7> 0\), the only acceptable solution is \(x=8\). Therefore, the correct answer is C.

We first need to evaluate \(f'(x)\). Using the product rule we obtain \begin{eqnarray*} f'(x)&=&x'\cdot e^{x^2+1}+x\cdot\left(e^{x^2+1}\right)'. \end{eqnarray*} We now use the chain rule to evaluate the derivative of \(e^{x^2+1}\). The derivative is \begin{eqnarray*} \left(e^{x^2+1}\right)'&=&e^{x^2+1}\cdot\left(x^2+1\right)'=e^{x^2+1}\cdot 2x. \end{eqnarray*} Therefore, the derivative of the function \(f\) satisfies \begin{eqnarray*} f'(x)&=&e^{x^2+1}+x\cdot e^{x^2+1}\cdot 2x\\ &=&e^{x^2+1}\left(1+2x^2\right). \end{eqnarray*} Substituting \(x=2\) gives us \(f'(2)=e^{2^2+1}\cdot\left(1+2\cdot 2^2\right)=9e^5\). The correct answer is A.

We will make the substitution \(u=3x+1\). With this substitution we have \(du=3\,dx\). The bounds in variable \(u\) are \(1\leq u\leq 13\). The integral satisfies \begin{eqnarray*} \int_0^4\frac1{3x+1}\,dx&=&\frac13\int_0^4\frac{3dx}{3x+1}=\frac13\int_1^{13}\frac{du}{u}\\ &=&\left.\frac13\ln|u|\right|_{u=1}^{u=13}=\frac13\left(\ln13-\ln 1\right) \\ &=&\frac{\ln13}3. \end{eqnarray*} The correct answer is C.

We will use the formula \(\ln\frac{\sqrt A}{B}=\ln\sqrt A-\ln B=\frac12\ln A-\ln B\) and the chain rule. The derivative \(\frac{dy}{dx}\) satisfies \begin{eqnarray*} \frac{dy}{dx}&=&\left(\frac12\ln(x-5)-\ln\left(4x^3+1\right)\right)'\\ &=&\frac12\left(\ln(x-5)\right)'-\left(\ln\left(4x^3+1\right)\right)'\\ &=&\frac12\cdot\frac1{x-5}\cdot\left(x-5\right)'-\frac1{4x^3+1}\cdot\left(4x^3+1\right)'\\ &=&\frac1{2(x-5)}-\frac{12x^2}{4x^3+1}. \end{eqnarray*} The correct answer is B.

Observe that the second derivative can be further factored into \(f''(x)=(3-x)(x-2)(x+2)\). The function is concave up whenever \(f''(x)> 0\). The zeroes of the function \(f''\) are \(-2\), \(2\), and \(3\). The function is positive on \((-\infty,-2)\) and (2,3). The correct answer is A.

The function \(f\) is increasing on the intervals on which \(f'\) is positive. The function \(f'\) is positive on \((-5,-1)\) and \((5,7)\). The correct answer is C.

We will break the interval of integration into three: \([0,2]\), \([2,3]\), and \([3,4]\). The function is positive on \([0,2]\) and negative on \([2,3]\) and \([3,4]\). Therefore, the integral over \([0,2]\) is the area under the graph. The integrals over \([2,3]\) and \([3,4]\) are the negative areas of the regions above the graph of the function and below the horizontal axis. The area under the graph is a right-angled triangle in the case of the interval \([0,2]\). Both sides have length \(2\), hence the area is \(\frac12\cdot2\cdot 2=2\). The area above the graph and below the interval \([2,3]\) is \(\frac12\), and the area between the graph and the interval \([3,4]\) is \(1\). Thus, the integral is \begin{eqnarray*} \int_0^4f(x)\,dx&=&\int_0^2f(x)\,dx+\int_2^3f(x)\,dx+\int_3^4f(x)\,dx\\ &=&2-\frac12-1=\frac12. \end{eqnarray*} The correct answer is A.

The velocity \(v(t)\) is the derivative of the position \(s(t)\). Therefore we have \begin{eqnarray*} s(t)&=&\int v(t)\,dt=\int\left(3t^2-4t+6\right)\,dt=t^3-2t^2+6t+C. \end{eqnarray*} The constant \(C\) is obtained from the initial condition \(s(1)=2\). Placing the value \(t=1\) in the equation \(s(t)\) gives us \(s(1)=1^3-2\cdot 1^2+6\cdot 1+C=5+C\). Therefore, \(C=-3\), and \[s(t)=t^3-2t^2+6t-3.\] The correct answer is C.

The linear approximation to \(f\) is \[L(x)=f(2)+f'(2)\cdot\left(x-2\right).\] The derivative of \(f\) is \(f'(x)=4x^3\). Therefore, \(f'(2)=32\), and the linearization is \[L(x)=16+32(x-2).\] The approximate value of \(f(2.01)\) is \(L(2.01)=16+32\cdot 0.01=16+0.32=16.32\). The correct answer is D.

The value \(f^{-1}(z)\) is the solution \(x\) to the equation \(f(x)=z\). The last equation is equivalent to \(\frac{4x-1}{2x+3}=z\). Multiplying both sides by \((2x+3)\) gives us \(4x-1=2xz+3z\), which after re-arranging the terms becomes \(x\cdot(4-2z)=1+3z\). Therefore the variable \(x\) in terms of \(z\) is \(x=\frac{1+3z}{4-2z}\). This means that \(f^{-1}(z)=\frac{1+3z}{4-2z}\). Substituting \(x\) instead of \(z\) gives us \[f^{-1}(x)=\frac{1+3x}{4-2x}.\] The correct answer is D.

We need to find \(x\) at which \(f''\) changes the sign from positive to negative, or from negative to positive. The first derivative of \(f\) is \(f'(x)=12x^2+18x+6\). The second derivative is \[f''(x)=24x+18.\] The function \(f''\) is negative for \(x< -\frac{18}{24} =-\frac34\) and positive for \(x> -\frac 34\). The correct answer is D.

The integral can be written as \begin{eqnarray*}\int_1^4\left(3\sqrt x+\frac{4}{x^2}\right)\,dx &=&\int_1^4\left(3 x^{\frac12}+4x^{-2}\right)\,dx\\ &=&\left.\left(3\cdot\frac23x^{\frac32}-4\cdot x^{-1}\right)\right|_{x=1}^{x=4}\\ &=&\left(3\cdot\frac234^{\frac32}-4\cdot 4^{-1}\right)- \left(3\cdot\frac231^{\frac32}-4\cdot 1^{-1}\right)\\ &=&17. \end{eqnarray*} The correct answer is E.

The sum is equal to \begin{eqnarray*} \sum_{i=1}^4i(4-i)&=&1\cdot (4-1)+ 2\cdot (4-2)+3\cdot (4-3)+4\cdot(4-4)\\ &=&3+4+3+0=10. \end{eqnarray*} The correct answer is C.

We will use the substitution \(u=x^3+2x\). With this substitution we have \(du=\left(3x^2+2\right)\,dx\). The integral becomes \begin{eqnarray*} \int \left(x^3+2x\right)^5\left(12x^2+8\right)\,dx&=&4\int \left(x^3+2x\right)^5\left(3x^2+2\right)\,dx\\ &=&4\int u^5\,du=4\cdot\frac16u^6+C\\ &=&\frac23\left(x^3+2x\right)^6+C. \end{eqnarray*} The correct answer is D.

The differential equation can be written in the form \(y^2\,dy=x^2\,dx\). Integrating both sides gives us \(\frac{y^3}3=\frac{x^3}3+C\). Multiplying by \(3\) yields \(y^3=x^3+3C\). Since \(C\) is a constant, \(3C\) is a also a constant and we can call it \(D\). The differential equation becomes \(y^3=x^3+D\). Solving for \(y\) gives us \(y=\sqrt[3]{x^3+D}\). From the initial condition \(y(0)=2\) we have \(\sqrt[3]{0^3+D}=2\). This implies that \(D=8\). Hence, \(y(x)=\sqrt[3]{x^3+8}\). The correct answer is E.

The elasticity of demand is given by \[\epsilon_D(p)=\frac{\frac{dx}{dp}}{\frac{x(p)}{p}}.\] The derivative of \(x(p)\) is \[x'(p)=\left(1000-3p^2\right)=-6p.\] Therefore, \[\epsilon_D(p)=-6p\cdot\frac{p}{x(p)}=-6p\cdot \frac{p}{1000-3p^2}.\] It remains to substitute \(p=10\). The elasticity of demand becomes \(\epsilon_D(10)=-6\cdot 10\cdot 10\cdot \frac1{700}=-\frac 67\). The correct answer is D.

From the conditions of the problem we have \(x-2y=16\). Therefore, we have \(x=16+2y\). The product of \(x\) and \(y\) is \(P=y(16+2y)\). We need to find the minimum of the function \(P(y)=y(16+2y)\). The function goes to \(+\infty\) if \(y\to\pm\infty\). Since the function is differentiable, the minimum is attained at the value \(y\) for which \(P'(y)=0\). The derivative is \[P'(y)=\left(16y+2y^2\right)'=16+4y.\] The solution to \(P'(y)=0\) is \(y=-4\). The value \(P(-4)\) is \[P(-4)=-4\cdot(16-2\cdot 4)=-32.\] The correct answer is A.

We will look for the values \(x\) for which \(f'(x)=0\). The derivative of \(f\) satisfies \[f'(x)=6x^2+6x-36=36(x^2+x-6)=36(x+3)(x-2).\] The equation \(f'(x)=0\) has two solutions. They are \(-3\) and \(2\). The function \(f'\) is positive for \(x< -3\) and negative for \(x\in(-3,+\infty)\). Therefore, \(f\) is increasing before \(-3\) and decreasing after \(-3\). This means that \(-3\) is a local maximum. The correct answer is A.

The function \(f\) is positive on \([-1,2]\). The approximate area is the same as the approximate integral \(\int_{-1}^2f(x)\,dx\). The width of each of the intervals of partition is \(1\). The endpoints are \(0\), \(1\), and \(2\). Therefore, the right-hand integral sum is \begin{eqnarray*} S&=&1\cdot f(0)+1\cdot f(1)+1\cdot f(2)=\left(5-0^2\right)+\left(5-1^2\right)+\left(5-2^2\right)\\ &=&5+4+1=10. \end{eqnarray*} The correct answer is B.

Since the second derivative is negative, the function is concave-down at \((2,7)\). The first derivative is decreasing at \(2\). Since the first derivative is \(0\), and it is decreasing, we conclude that the first derivative must be positive before \(2\) and negative after. Thus, \((2,7)\) is a local maximum. The correct answer is A.

The average value is \begin{eqnarray*} A&=&\frac1{2-0}\int_0^2f(x)\,dx=\frac12\int_0^2\left(4x-x^2\right)\,dx\\ &=&\left.\frac12\left(2x^2-\frac13x^3\right)\right|_{x=0}^{x=2}\\ &=&\frac12\left(2\cdot 2^2-\frac132^3\right)-\frac12\left(2\cdot 0^2-\frac13\cdot 0^3\right)\\ &=&\frac83. \end{eqnarray*} The correct answer is E.