Question
If \(f'(x)\) and \( g'( x)\) exist and \(f'(x)>g'(x)\) for all real \(x\), then the graph of \(y = f(x) \) and the graph of \( y= g(x )\)
(A) intersect exactly once.
(B) intersect no more than once.
(C) do not intersect.
(D) could intersect more than once.
(E) have a common tangent at each point of intersection.
▶️ Answer/Explanation
Hint:
\( f'(x) > g'(x) \) for all \( x \), \( f(x) – g(x) \) is increasing (derivative \( f'(x) – g'(x) > 0 \). A strictly increasing function crosses zero at most once.
If \( y \) is a function of \( x \) such that \( y’ > 0 \) for all \( x \) and \( y” < 0 \) for all \( x \), which of the following could be part of the graph of \( y = f(x) \)?
▶️ Answer/Explanation
Hint:
\( y’ > 0 \) means the function is strictly increasing for all \( x \),
\( y” < 0 \) means the function is concave down, so the graph curves downward without inflection points.
Question
Let \( f \) be a function that is continuous on the closed interval \([-2, 3]\) such that \( f'(0) \) does not exist, \( f'(2) = 0 \), and \( f”(x) < 0 \) for all \( x \) except \( x = 0 \). Which of the following could be the graph of \( f \)?
▶️Answer/Explanation
Ans:E
Hint:
Graphs A and B contradict \(f'<0\). Graph C contradicts \(f'(0)\) does not exist. Graph D contradicts continuity on the interval [−2,3]. Graph E meets all given conditions.
Question
Which of the following pairs of graphs could represent the graph of a function and the graph of its derivative?
(A) I only (B) II only (C) III only (D) I and III (E) II and III
▶️Answer/Explanation
Ans:D
Hint:
II does not work since the slope of f at x = 0 is not equal to f ′(0). Both I and III could work. For example,\(f ′( x) =e^x\) in I and \(f (x)=sinx\) in III
Question
If \( x + 7y = 29 \) is an equation of the line normal to the graph of \( f \) at the point \( (1, 4) \), then \( f'(1) = \)
(A) 7
(B) \(\frac{1}{7}\)
(C) \(-\frac{1}{7}\)
(D) \(-\frac{7}{29}\)
(E) -7
▶️Answer/Explanation
Ans:A
Hint:
The normal line is perpendicular to the tangent line. \( y = -\frac{1}{7}x + \frac{29}{7} \). Slope of the normal line: \( -\frac{1}{7} \). Slope of the tangent line (negative reciprocal): \( -\frac{1}{-\frac{1}{7}} = 7 \). Thus, \( f'(1) = 7 \).