AP Calculus BC 1.11 Defining Continuity at a Point Exam Style Questions - FRQ
Question:
Let f be a function defined by \(\left\{\begin{matrix}
1-2 sin x & for x \leq 0 & & \\e^{-4x}
&for x > 0. & &
\end{matrix}\right.\)
(a) Show that f is continuous at x = 0.
(b) For x ≠ 0, express f'(x) as a piecewise-defined function. Find the value of x for which f'(x) = -3.
(c) Find the average value of f on the interval [-1, 1] .
Answer/Explanation
Ans:
(a)
To be continuous
i) f(0) = 1
ii) \(\lim_{x\rightarrow 0} 1 -2 sin x = \lim_{x\rightarrow 0}e^{-4x}\)
1 = 1 ∴ \(\lim_{x\rightarrow 0^{-}} f = \lim_{x\rightarrow 0^{+}}f\)
∴ \(\lim_{x\rightarrow 0} f = 1\)
iii) f(0) = \(\lim_{x\rightarrow 0} f = 1\)
∴ f is continuous for all values of x.
(b)
\(f'(x)=\left\{\begin{matrix}
-2 cos x &, &x<0 & \\-4e^{-4x}
&, &x>0 &
\end{matrix}\right.\)
\({f}'(x) = 3\)
Since -2 cos x oscillates between -2 and 2 there will be no such value in this function such that f'(x) = -3
but \({f}'(x) =- 3\)
-4e -4x = -3
e -4x = 3/4
-4x = In (3/4) \(\therefore = \frac{-1}{4}In\left ( \frac{3}{4} \right )\therefore f’\left ( \frac{-1}{4}In\left ( \frac{3}{4} \right ) \right )= -3\)
(c)
\(f_{avg} = \frac{1}{1+1}\int_{-1}^{1}f(x)dx\)
\(= \frac{1}{2}\left [ \int_{-1}^{0} 1-2 sinx dx + \int_{0}^{1}e^{-4x}dx\right ]\)
\(= \frac{1}{2}\left [ [x+2cosx]_{-1}^{0}+[\frac{-1}{4}e^{-4x}]_{0}^{1} \right ]\)
\(= \frac{1}{2}\left [ 2+1-2cos(-1)\frac{-e^{-4}}{4} +\frac{1}{4}\right ]\)
\(=\frac{-\left [ 3-2cos(-1) \right ]+\left [ \frac{-1}{4e^{4}} +\frac{1}{4}\right ]}{2}\)