Question
(a) Topic-2.1-Defining Average and Instantaneous Rates of Change at a Point
(b) Topic-2.10-Finding the Derivatives of Tangent Cotangent Secant and or Cosecant Functions
(c) Topic-5.6-Determining Concavity of Functions over Their Domains
(d) Topic-4.7-Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms
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
\(\textbf{5(a)} \quad \text{The average rate of change of } f \text{ on the interval } 0 \leq x \leq \pi \text{ is}\)
\(\quad \frac{f(\pi) – f(0)}{\pi – 0} = \frac{-e^{\pi} – 1}{\pi}.\)
\(\textbf{5(b)} \quad f'(x) = e^x \cos x – e^x \sin x\)
\(\quad f’\left(\frac{3\pi}{2}\right) = e^{3\pi/2} \cos\left(\frac{3\pi}{2}\right) – e^{3\pi/2} \sin\left(\frac{3\pi}{2}\right) = e^{3\pi/2}\)
\(\quad \text{The slope of the line tangent to the graph of } f \text{ at } x = \frac{3\pi}{2} \text{ is } e^{3\pi/2}.\)
\(\textbf{5(c)} \quad f'(x) = 0 \implies \cos x – \sin x = 0 \implies x = \frac{\pi}{4}, \, x = \frac{5\pi}{4}\)
\(\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
0 & 1 \\
\frac{\pi}{4} & \frac{1}{\sqrt{2}} e^{\pi/4} \\
\frac{5\pi}{4} & -\frac{1}{\sqrt{2}} e^{5\pi/4} \\
2\pi & e^{2\pi} \\
\hline
\end{array}\)
\(\quad \text{The absolute minimum value of } f \text{ on } 0 \leq x \leq 2\pi \text{ is } -\frac{1}{\sqrt{2}} e^{5\pi/4}.\)
\(\textbf{5(d)} \quad \lim_{x \to \pi/2} f(x) = 0\)
\(\quad \text{Because } g \text{ is differentiable, } g \text{ is continuous.}\)
\(\quad \lim_{x \to \pi/2} g(x) = g\left(\frac{\pi}{2}\right) = 0\)
\(\quad \text{By L’Hospital’s Rule,}\)
\(\quad \lim_{x \to \pi/2} g(x) = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{-e^{\pi/2}}{2}.\)