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Question 1. Does the following graph have a point of inflection?
Step 1: A point of inflection occurs where the concavity changes. Step 2: Look at the graph carefully. To the left of \(x=0\), the graph is concave down. Step 3: To the right of \(x=0\), the graph is concave up. Step 4: Since the graph changes from concave down to concave up, there is a point of inflection. Answer: True
2.
Question 2. The function \(f(x)=x^3+2\) is increasing on the interval \((0,\infty)\).
Step 1: Find the derivative of \(f(x)=x^3+2\). Step 2: \(f'(x)=3x^2\). Step 3: A function is increasing where its derivative is positive. Step 4: On the interval \((0,\infty)\), we have \(x>0\), so \(3x^2>0\). Step 5: Therefore, \(f(x)=x^3+2\) is increasing on \((0,\infty)\). Answer: True
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Question 3. For which value of \(x\) does \(f(x)=\frac{x^4-x^2}{x+1}\) have a point of inflection?
Step 1: Factor the numerator: \(x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1)\). Step 2: For \(x\ne -1\), simplify \(\frac{x^4-x^2}{x+1}=x^2(x-1)=x^3-x^2\). Step 3: Differentiate once: \(f'(x)=3x^2-2x\). Step 4: Differentiate again: \(f''(x)=6x-2\). Step 5: Set the second derivative equal to zero: \(6x-2=0\). Step 6: Solve: \(6x=2\), so \(x=\frac{1}{3}\). Step 7: Note that \(x=-1\) is not in the domain of the original function, so it cannot be the point of inflection. Answer: \(x=\frac{1}{3}\)
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Question 4. How many relative minimums does \(\sin x\) have on the interval \((-2\pi,2\pi)\)?
Step 1: Let \(f(x)=\sin x\). Step 2: Find the first derivative: \(f'(x)=\cos x\). Step 3: Critical points occur when \(\cos x=0\). Step 4: On \((-2\pi,2\pi)\), the critical points are \(x=-\frac{3\pi}{2}\), \(x=-\frac{\pi}{2}\), \(x=\frac{\pi}{2}\), and \(x=\frac{3\pi}{2}\). Step 5: Use the second derivative test. Since \(f''(x)=-\sin x\), a relative minimum occurs when \(f''(x)>0\). Step 6: At \(x=-\frac{\pi}{2}\), \(f''(x)=-\sin\left(-\frac{\pi}{2}\right)=1>0\), so this is a relative minimum. Step 7: At \(x=\frac{3\pi}{2}\), \(f''(x)=-\sin\left(\frac{3\pi}{2}\right)=1>0\), so this is also a relative minimum. Step 8: Therefore, there are \(2\) relative minimums. Answer: 2
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Question 5. The number of points of inflection of \(f(x)=x^{\frac{1}{5}}\) is
Step 1: Find the first derivative of \(f(x)=x^{\frac{1}{5}}\): \(f'(x)=\frac{1}{5}x^{-\frac{4}{5}}\). Step 2: Find the second derivative: \(f''(x)=-\frac{4}{25}x^{-\frac{9}{5}}\). Step 3: Rewrite the second derivative as \(f''(x)=-\frac{4}{25x^{\frac{9}{5}}}\). Step 4: The second derivative is undefined at \(x=0\). Step 5: The original function \(f(x)=x^{\frac{1}{5}}\) is continuous at \(x=0\), so \(x=0\) can be a possible inflection point. Step 6: Test the sign of \(f''(x)\) on both sides of \(0\). Step 7: If \(x<0\), \(x^{\frac{9}{5}}0\), meaning concave up. Step 8: If \(x>0\), \(x^{\frac{9}{5}}>0\), so \(f''(x)<0\), meaning concave down. Step 9: Since concavity changes at \(x=0\), there is one point of inflection. Answer: 1
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Question 6. The function \(f(x)=x^4+8x^3\) has
Step 1: Find the first derivative: \(f'(x)=4x^3+24x^2\). Step 2: Factor the derivative: \(f'(x)=4x^2(x+6)\). Step 3: Set \(f'(x)=0\). This gives \(x=0\) and \(x=-6\). Step 4: Find the second derivative: \(f''(x)=12x^2+48x\). Step 5: Evaluate at \(x=-6\): \(f''(-6)=12(-6)^2+48(-6)=432-288=144>0\). Step 6: Since \(f''(-6)>0\), there is a local minimum at \(x=-6\). Step 7: Evaluate at \(x=0\): \(f''(0)=0\), so the second derivative test is inconclusive. Step 8: Because \(f'(x)=4x^2(x+6)\), the sign of \(f'(x)\) does not change at \(x=0\), so there is no local maximum or local minimum there. Answer: a local min at \(x=-6\)
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Question 7. A fence is to be used to entirely enclose a rectangular plot of land with an area of \(1600\text{ ft}^2\). It can be shown that a \(40\text{ ft}\) by \(40\text{ ft}\) rectangle will require the least amount of fencing to enclose this plot of land.
Step 1: Let the rectangle have length \(l\), width \(w\), area \(A\), and perimeter \(P\). Step 2: The area is fixed: \(A=lw=1600\). Step 3: Solve for \(l\): \(l=\frac{1600}{w}\). Step 4: The amount of fencing is the perimeter: \(P=2l+2w\). Step 5: Substitute \(l=\frac{1600}{w}\): \(P=2\left(\frac{1600}{w}\right)+2w=\frac{3200}{w}+2w\). Step 6: Differentiate: \(P'=-\frac{3200}{w^2}+2\). Step 7: Set \(P'=0\): \(2-\frac{3200}{w^2}=0\). Step 8: Solve: \(2w^2=3200\), so \(w^2=1600\), giving \(w=40\). Step 9: Find \(l\): \(l=\frac{1600}{40}=40\). Step 10: Therefore, the minimum amount of fencing occurs with a \(40\text{ ft}\) by \(40\text{ ft}\) rectangle. Answer: True
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Question 8. A rectangular sheet of cardboard with dimensions \(5\text{ cm}\) by \(10\text{ cm}\) is used to make an open box by cutting squares of equal size from the four corners and folding up the sides. The maximum volume that can be obtained this way is
Step 1: Let \(x\) be the side length of each square cut from the corners. Step 2: After cutting and folding, the box has height \(x\), length \(10-2x\), and width \(5-2x\). Step 3: Write the volume function: \(V=x(10-2x)(5-2x)\). Step 4: Expand: \(V=x(50-30x+4x^2)=4x^3-30x^2+50x\). Step 5: Differentiate: \(V'=12x^2-60x+50\). Step 6: Set \(V'=0\): \(12x^2-60x+50=0\). Step 7: Use the quadratic formula: \(x=\frac{60\pm\sqrt{60^2-4(12)(50)}}{2(12)}\). Step 8: Simplify: \(x=\frac{60\pm\sqrt{1200}}{24}\), giving approximately \(x=1.06\) or \(x=3.94\). Step 9: Since the cardboard width is \(5\), the cut size must satisfy \(0<x<2.5\), so \(x=3.94\) is not possible. Step 10: Use \(x\approx1.06\). Step 11: Calculate the volume: \(V=(1.06)(10-2(1.06))(5-2(1.06))\approx24.06\text{ cm}^3\). Step 12: Since \(24.06\) is between \(20\) and \(30\), the correct choice is between \(20\) and \(30\text{ cm}^3\). Answer: between \(20\) and \(30\text{ cm}^3\)
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