(a)
The average rate of change of f on the interval 0 ≤ x ≤ π is
[tex]\displaystyle
f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi} [/tex]
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(b)
[tex]f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\
f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\
f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}[/tex]
The slope of the tangent line is [tex]e^{3\pi/2}[/tex].
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(c)
The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.
Solving f'(x) = 0
[tex]f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\
0 = e^x \big( \cos(x) - \sin(x)\big)[/tex]
Use zero factor property to solve.
[tex]e^x \ \textgreater \ 0\forall x \in \mathbb{R}[/tex] so that factor will not generate solutions.
Set cos(x) - sin(x) = 0
[tex]\cos (x) - \sin (x) = 0 \\
\cos(x) = \sin(x) [/tex]
cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):
[tex]\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\
x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi][/tex]
We check the values of f at the end points and these two critical numbers.
[tex]f(0) = e^1 \cos(0) = 1[/tex]
[tex]\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4} \frac{\sqrt{2}}{2}[/tex]
[tex]\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4} \frac{-\sqrt{2}}{2} = -e^{\pi/4} \frac{\sqrt{2}}{2}[/tex]
[tex]f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}[/tex]
There is only one negative number.
The absolute minimum value of f on the interval 0 ≤ x ≤ 2π is
[tex]-e^{5\pi/4} \sqrt{2}/2[/tex]
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(d)
The function f is a continuous function as it is a product of two continuous functions. Therefore, [tex]\lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0[/tex]
g is a differentiable function; therefore, it is a continuous function, which tells us [tex]\lim_{x \to \pi/2} g(x) = g(\pi/2) = 0[/tex].
When we observe the limit [tex]\displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}[/tex], the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.
[tex]\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)} [/tex]
[tex]f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\
g'(\pi/2) = 2[/tex]
thus
[tex]\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}[/tex]
