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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. 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 when the concavity of the function changes. Step 2: The condition \(f''(c)=0\) only tells us that the second derivative is zero at \(x=c\). Step 3: This condition alone does not guarantee that concavity changes from concave up to concave down, or from concave down to concave up. Step 4: We must also check the sign of \(f''(x)\) on both sides of \(c\). Step 5: Therefore, \(f''(c)=0\) does not necessarily mean \(x=c\) is a point of inflection. Answer: False
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Question 3. The function \(f(x)=\sin x\) is increasing on one of the following intervals.
Step 1: To find where \(f(x)=\sin x\) is increasing, 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 the interval \(0<x0\), so \(\sin x\) is increasing. Step 5: On \(\frac{\pi}{2}<x<\pi\), \(\cos x<0\), so the function is decreasing. Step 6: Therefore, the correct interval is \(\left(0,\frac{\pi}{2}\right)\). Answer: \(\left(0,\frac{\pi}{2}\right)\)
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Question 4. The function \(f(x)=2x^2-8x+1\) on the domain \([0,5]\) has an absolute maximum of
Step 1: Find the derivative: \(f'(x)=4x-8\). Step 2: Set the derivative equal to zero to find critical points: \(4x-8=0\). Step 3: Solve: \(4x=8\), so \(x=2\). Step 4: Evaluate the function at the critical point and the endpoints of \([0,5]\). Step 5: \(f(0)=2(0)^2-8(0)+1=1\). Step 6: \(f(2)=2(2)^2-8(2)+1=8-16+1=-7\). Step 7: \(f(5)=2(5)^2-8(5)+1=50-40+1=11\). Step 8: Compare the values \(1\), \(-7\), and \(11\). The largest value is \(11\). Answer: 11
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Question 5. \(f(x)=\ln(x+1)\) has no relative maximum.
Step 1: Find the derivative: \(f'(x)=\frac{1}{x+1}\). Step 2: A relative maximum can occur at a critical point where \(f'(x)=0\) or where \(f'(x)\) does not exist within the domain. Step 3: Set \(f'(x)=0\): \(\frac{1}{x+1}=0\). Step 4: This equation has no solution because a fraction with numerator \(1\) cannot equal \(0\). 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
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Question 6. A critical point of the function \(f(x)=\sqrt{x+3}\) is
Step 1: Find the domain of \(f(x)=\sqrt{x+3}\). We need \(x+3\ge0\), so \(x\ge-3\). Step 2: Differentiate: \(f'(x)=\frac{1}{2\sqrt{x+3}}\). Step 3: A critical point occurs where \(f'(x)=0\) or where \(f'(x)\) does not exist, as long as the point is in the domain of the original function. Step 4: The derivative \(f'(x)=\frac{1}{2\sqrt{x+3}}\) is never equal to \(0\). Step 5: The derivative is undefined when \(\sqrt{x+3}=0\), so \(x+3=0\), giving \(x=-3\). Step 6: Since \(x=-3\) is in the domain of \(f(x)\), it is a critical point. Answer: \(-3\)
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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 \(40\), we have \(x+y=40\). Step 3: Solve for one variable: \(x=40-y\). Step 4: The product is \(P=xy\). Substitute \(x=40-y\): \(P=(40-y)y=40y-y^2\). Step 5: This quadratic opens downward, so its critical point gives a maximum product, not a minimum product. Step 6: For positive integers with a fixed sum, the smallest product occurs when the numbers are as far apart as possible. Step 7: The farthest-apart positive integers that add to \(40\) are \(1\) and \(39\). Step 8: Check: \(1+39=40\), and the product is \(1\cdot39=39\), which is smaller than the other choices. Answer: 1, 39
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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 \(P=2l+2w=20\). Step 3: Solve for \(l\): \(2l=20-2w\), so \(l=10-w\). Step 4: The area is \(A=lw\). Substitute \(l=10-w\): \(A=(10-w)w=10w-w^2\). Step 5: Differentiate the area function: \(A'=10-2w\). Step 6: Set the derivative equal to zero: \(10-2w=0\). Step 7: Solve: \(2w=10\), so \(w=5\). Step 8: Find \(l\): \(l=10-w=10-5=5\). Step 9: The rectangle with maximum area is a square with side lengths \(5\) and \(5\). Answer: True
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