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Question 1. If \(f\) is continuous everywhere, \(f'(3)=0\) and \(f''(3)=4\), then there must be a local minimum at \(x=3\).
Step 1: Since \(f'(3)=0\), \(x=3\) is a critical point. Step 2: Use the second derivative test to classify the critical point. Step 3: Since \(f''(3)=4>0\), the graph is concave up at \(x=3\). Step 4: A critical point where the graph is concave up is a local minimum. Step 5: Therefore, there must be a local minimum at \(x=3\). Answer: True
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Question 2. If \(f''(c)=0\), then \(x=c\) must be a point of inflection.
Step 1: A point of inflection occurs where the concavity of the graph changes. Step 2: The equation \(f''(c)=0\) only tells us that the second derivative is zero at \(x=c\). Step 3: A zero second derivative does not always mean that concavity changes. Step 4: To confirm a point of inflection, we must check the sign of \(f''(x)\) on both sides of \(c\). Step 5: Since concavity change is not guaranteed, the statement is false. Answer: False
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Question 3. For which \(x\)-value(s) is \(f(x)=x^4-4x^2+2\) concave down? i) \(x=-1\), ii) \(x=0\), iii) \(x=1\).
Step 1: Concavity is determined by the second derivative. Step 2: Differentiate \(f(x)=x^4-4x^2+2\): \(f'(x)=4x^3-8x\). Step 3: Differentiate again: \(f''(x)=12x^2-8\). Step 4: A function is concave down where \(f''(x)0\), so it is concave up. Step 6: Test \(x=0\): \(f''(0)=12(0)^2-8=-80\), so it is concave up. Step 8: Only \(x=0\) gives concave down. Answer: ii)
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Question 6. 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 where \(\cos x=0\). 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: Find the second derivative: \(f''(x)=-\sin x\). Step 6: A relative minimum occurs where \(f''(x)>0\). Step 7: At \(x=-\frac{\pi}{2}\), \(f''(x)=-\sin\left(-\frac{\pi}{2}\right)=1>0\), so there is a relative minimum. Step 8: At \(x=\frac{3\pi}{2}\), \(f''(x)=-\sin\left(\frac{3\pi}{2}\right)=1>0\), so there is another relative minimum. Step 9: Therefore, there are \(2\) relative minimums. Answer: 2
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Question 5. What is the absolute minimum of \(f(x)=\sqrt{x-4}-3\)?
Step 1: Determine the domain of \(f(x)=\sqrt{x-4}-3\). Step 2: Since the expression under the square root must be nonnegative, \(x-4\ge0\), so \(x\ge4\). Step 3: The square root \(\sqrt{x-4}\) is always greater than or equal to \(0\). Step 4: Therefore, \(\sqrt{x-4}-3\ge -3\). Step 5: The smallest value occurs when \(x=4\). Step 6: Evaluate: \(f(4)=\sqrt{4-4}-3=0-3=-3\). Answer: \(y=-3\)
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Question 4. The number of points of inflection of \(f(x)=x^{\frac{1}{5}}\) is
Step 1: Find the first derivative: \(f'(x)=\frac{1}{5}x^{-\frac{4}{5}}\). Step 2: Find the second derivative: \(f''(x)=-\frac{4}{25}x^{-\frac{9}{5}}\). Step 3: Rewrite this as \(f''(x)=-\frac{4}{25x^{\frac{9}{5}}}\). Step 4: The second derivative is undefined at \(x=0\). Step 5: The original function \(x^{\frac{1}{5}}\) is continuous at \(x=0\), so \(x=0\) is a possible inflection point. Step 6: Test concavity on both sides of \(0\). Step 7: When \(x<0\), \(x^{\frac{9}{5}}0\), meaning concave up. Step 8: When \(x>0\), \(x^{\frac{9}{5}}>0\), so \(f''(x)<0\), meaning concave down. Step 9: Since concavity changes at \(x=0\), there is exactly one point of inflection. Answer: 1
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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 \(l\) be the length and \(w\) be the width of the rectangle. Step 2: The area is fixed: \(lw=1600\). 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 \(l=\frac{1600}{40}=40\). Step 10: Therefore, the rectangle requiring the least fencing is \(40\text{ ft}\) by \(40\text{ ft}\). Answer: True
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Question 8. A rectangle has a perimeter of \(20\). If the area of the rectangle is to be maximized, the length of each side should be \(5\).
Step 1: Let the rectangle have length \(l\) and width \(w\). Step 2: The perimeter is \(2l+2w=20\). Step 3: Solve for \(l\): \(2l=20-2w\), so \(l=10-w\). Step 4: The area is \(A=lw\). Step 5: Substitute \(l=10-w\): \(A=(10-w)w=10w-w^2\). Step 6: Differentiate: \(A'=10-2w\). Step 7: Set \(A'=0\): \(10-2w=0\). Step 8: Solve: \(w=5\). Step 9: Find \(l=10-w=10-5=5\). Step 10: The maximum area occurs when the rectangle is a square with side lengths \(5\) and \(5\). Answer: True
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