AP Calculus AB 2.2 Defining the Derivative of a Function and Using Derivative Notation - MCQs - Exam Style Questions
No-Calc Question
\(\displaystyle \lim_{h\to 0}\frac{e^{-(1+h)}-e^{-1}}{h}\) is
(A) \(-\tfrac{1}{e}\)
(B) \(0\)
(C) \(\tfrac{1}{e}\)
(D) nonexistent
(B) \(0\)
(C) \(\tfrac{1}{e}\)
(D) nonexistent
▶️ Answer/Explanation
Recognize a derivative: \(\dfrac{f(1+h)-f(1)}{h}\) with \(f(x)=e^{-x}\).
Then the limit equals \(f'(1)=-e^{-1}=-\dfrac{1}{e}\).
(Alternatively, factor \(e^{-1}\) and use \(\lim_{h\to 0}\dfrac{e^{-h}-1}{h}=-1\).)
✅ Answer: (A)
Then the limit equals \(f'(1)=-e^{-1}=-\dfrac{1}{e}\).
(Alternatively, factor \(e^{-1}\) and use \(\lim_{h\to 0}\dfrac{e^{-h}-1}{h}=-1\).)
✅ Answer: (A)
No-Calc Question
The function \(f\) satisfies the properties shown. What is the slope of the line tangent to the graph of \(f\) at \(x=-1\) ?
\(f(0)=0,\qquad \displaystyle\lim_{x\to -1}f(x)=\dfrac{2}{5},\qquad \displaystyle\lim_{x\to -1}\dfrac{f(x)-f(-1)}{x+1}=-\dfrac{3}{5}.\)
(A) \(-\dfrac{3}{5}\)
(B) \(-\dfrac{2}{5}\)
(C) \(\dfrac{2}{5}\)
(D) \(1\)
▶️ Answer/Explanation
By definition, \(f'(-1)=\displaystyle\lim_{x\to -1}\dfrac{f(x)-f(-1)}{x+1}\).
The given limit equals \(-\dfrac{3}{5}\).
✅ Answer: (A)