1.
Question 1. The function \(f(x)=\sin x\) is increasing on one of the following intervals:
Step 1: Find the derivative: \(f'(x)=\cos x\). Step 2: A function is increasing where its derivative is positive. Step 3: We need \(\cos x>0\). Step 4: On \(\left(0,\frac{\pi}{2}\right)\), \(\cos x>0\). Step 5: On the other listed intervals, \(\cos x<0\), so \(\sin x\) is decreasing there. Answer: \(\left(0,\frac{\pi}{2}\right)\)
2.
Question 2. For the continuous function \(f(x)\), if \(f'(-4)\) does not exist, then \(x=-4\) is a critical point.
Step 1: A critical point occurs where the function is defined and the derivative is \(0\) or does not exist. Step 2: The function is continuous, so \(f(-4)\) exists. Step 3: The derivative \(f'(-4)\) does not exist. Step 4: Since the function exists and the derivative does not exist at \(x=-4\), it is a critical point. Answer: True
3.
Question 3. For which values of \(x\) does \(f(x)=\frac{x^4-x^2}{x+1}\) have a point of inflection? Choose all answers that apply.
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: Since the original function is undefined at \(x=-1\), \(x=-1\) cannot be an inflection point. Step 4: Differentiate: \(f'(x)=3x^2-2x\). Step 5: Differentiate again: \(f''(x)=6x-2\). Step 6: Set \(f''(x)=0\): \(6x-2=0\). Step 7: Solve: \(x=\frac{1}{3}\). Answer: \(x=\frac{1}{3}\)
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Question 4. \(f(x)=\sqrt{x}\) has an absolute minimum at \(x=0\).
Step 1: The domain of \(f(x)=\sqrt{x}\) is \(x\ge0\). Step 2: For all \(x\ge0\), \(\sqrt{x}\ge0\). Step 3: The smallest possible value is \(0\). Step 4: This occurs at \(x=0\), since \(f(0)=0\). Answer: True
5.
Question 5. \(f(x)=\ln(x+1)\) has no relative maximum.
Step 1: Differentiate: \(f'(x)=\frac{1}{x+1}\). Step 2: A relative maximum can occur where \(f'(x)=0\) or where \(f'(x)\) does not exist in the domain. Step 3: Solve \(\frac{1}{x+1}=0\). Step 4: This has no solution. Step 5: The derivative is undefined at \(x=-1\), but \(x=-1\) is not in the domain of \(\ln(x+1)\). Step 6: Therefore, there is no relative maximum. Answer: True
6.
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: \(f'(x)=4x^2(x+6)\). Step 3: Set \(f'(x)=0\), giving \(x=0\) and \(x=-6\). Step 4: Find the second derivative: \(f''(x)=12x^2+48x\). Step 5: Evaluate \(f''(-6)=12(-6)^2+48(-6)=144>0\), so there is a local minimum at \(x=-6\). Step 6: Evaluate \(f''(0)=0\), so the second derivative test is inconclusive at \(x=0\). Step 7: Since \(f'(x)=4x^2(x+6)\) does not change sign at \(x=0\), there is no local maximum or minimum there. Answer: A local min at \(x=-6\).
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Question 7. Which of the following two positive integers, whole numbers, both add to \(40\) and have the smallest product?
Step 1: Let the two positive integers be \(x\) and \(y\). Step 2: Since they add to \(40\), \(x+y=40\). Step 3: The product is smallest when the two positive integers are as far apart as possible. Step 4: The smallest positive integer is \(1\), so the other integer is \(39\). Step 5: Check: \(1+39=40\) and \(1\cdot39=39\), which is smaller than the other choices. Answer: 1, 39
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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: The box height is \(x\), length is \(10-2x\), and width is \(5-2x\). Step 3: Write the volume: \(V=x(10-2x)(5-2x)\). Step 4: Expand: \(V=4x^3-30x^2+50x\). Step 5: Differentiate: \(V'=12x^2-60x+50\). Step 6: Set \(V'=0\): \(12x^2-60x+50=0\). Step 7: Use the quadratic formula: \(x=\frac{60\pm\sqrt{1200}}{24}\), so \(x\approx3.94\) or \(x\approx1.06\). Step 8: Since \(0<x<2.5\), \(x=3.94\) is impossible. Step 9: Use \(x=1.06\). Step 10: \(V\approx(1.06)(10-2(1.06))(5-2(1.06))\approx24.06\text{ cm}^3\). Step 11: \(24.06\) is between \(20\) and \(30\). Answer: Between \(20\) and \(30\text{ cm}^3\)
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