Question:
Let f be the function defined by f(x) = ex cos x.
(a) Find the average rate of change of f on the interval 0 ≤ x ≤ π .
(b) What is the slope of the line tangent to the graph of f at \(x = \frac{3\pi }{2}?\)
(c) Find the absolute minimum value of f on the interval 0 ≤ x ≤ 2π. Justify your answer.
(d) Let g be a differentiable function such that \(g\left ( \frac{\pi }{2} \right )=0.\) The graph of g’, the derivative of g, is shown below. Find the value of \(\lim_{x\rightarrow \pi /2}\frac{f(x)}{g(x)}\) or state that it does not exist. Justify your answer.
▶️Answer/Explanation
Ans:
(a)
\(\frac{\int_{0}^{r}f'(x)dx}{r}=\frac{f(r)-f(0)}{r}=\frac{e^{r}cosr-e^{0}cos 0}{r}\)
\(\frac{e^{r}(-1)-1(1)}{r}=\frac{-e^{r}-1}{r}\)
(b)
\(\frac{df}{dx}=e^{x}cos x + e^{x}(-sinx)\)
\(=e^{x}(cos x -sinx)\)
\(=e^{3r/2}(0-(-1))=e^{3r/2}\)
(c)
from part B
\(f'(x)=e^{x}(cos x – sin x)= 0 \rightarrow e^{x}=0\) or cox = sin x \(x = \frac{\pi }{4},\frac{5\pi}{4}\)
Minima either have a derivative at 0, undefined, or are boundary points.
So possible values : x = 0, x = 2π, x = π/4 x = 5π/4
f(x) = 1 f(x) = e2π \(f(x)=e^{\pi /4}\left ( \frac{\sqrt{2}}{2} \right )\) \(f(x)=e^{5\pi /4}\left ( -\frac{\sqrt{2}}{2} \right )\)
Since \(e^{5\pi /4}\left ( -\frac{\sqrt{2}}{2} \right )\) is the only negative value of f(x), the absolute minimum of f(x) on 0 ≤ x ≤ 2π is \(-\frac{\sqrt{2}}{2}e^{5\pi /4}\) .
(d)
Note \(\lim_{x\rightarrow \frac{\pi }{2}}(g(x))=0\) since g(x) is differentiable (and thus continuous), and \(g\left ( \frac{\pi }{2} \right )=0.\) Also \(f\left ( \frac{\pi }{2} \right )=e^{\pi /2}\left ( cos\left ( \frac{\pi }{2} \right ) \right )=e^{\pi /2}(0)=0,\) and f(x) is continuous, So \(\lim_{x\rightarrow \frac{\pi }{2}}f(x)=0.\) So by L’ Hopital’s rule,
\(\lim_{x\rightarrow \frac{\pi }{2}}\left ( \frac{f(x)}{g(x)} \right )=\lim_{x\rightarrow \frac{2 }{2}}\left ( \frac{f'(x)}{g'(x)} \right )=\frac{e^{\pi /2}\left ( cos\frac{\pi }{2}-sin\frac{\pi }{2} \right )}{2}=\frac{-e^{\pi /2}}{2}\)