Question

Let f be the function defined by f(x) = e^(x) cos x.

Answer 1

(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

   $\displaystyle f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}$

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(b)

$f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\ f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\ f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}$

The slope of the tangent line is $e^{3\pi/2}$.

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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

$f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\ 0 = e^x \big( \cos(x) - \sin(x)\big)$

Use zero factor property to solve.

$e^x \ \textgreater \ 0\forall x \in \mathbb{R}$ so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

$\cos (x) - \sin (x) = 0 \\ \cos(x) = \sin(x)$

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):

$\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\ x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]$

We check the values of f at the end points and these two critical numbers.

$f(0) = e^1 \cos(0) = 1$

$\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4} \frac{\sqrt{2}}{2}$

$\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4} \frac{-\sqrt{2}}{2} = -e^{\pi/4} \frac{\sqrt{2}}{2}$

$f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}$

There is only one negative number.
The absolute minimum value of f on the interval 0 ≤ x ≤ 2π is
$-e^{5\pi/4} \sqrt{2}/2$

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(d)

The function f is a continuous function as it is a product of two continuous functions. Therefore, $\lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0$

g is a differentiable function; therefore, it is a continuous function, which tells us $\lim_{x \to \pi/2} g(x) = g(\pi/2) = 0$.

When we observe the limit  $\displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}$, the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)}$

$f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\ g'(\pi/2) = 2$

thus

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}$