AP Calculus AB Exam Question:

Let $f(x) = e^x \cos(x)$.

a) Find the equation of the derivative of $f(x)$.

b) Find the second derivative of $f(x)$.

c) Find the point(s) of inflection for $f(x)$.

d) Find the intervals of concavity for $f(x)$.

e) Determine the intervals where $f(x)$ is increasing or decreasing.

Answer and Step-by-Step Explanation:

a) The derivative of $f(x) = e^x \cos(x)$ can be found by applying the product rule:

Recall the product rule states that the derivative of the product of two functions $u(x)$ and $v(x)$ is given by: $(u*v)' = u'v + uv'$.

Using the product rule, let's differentiate $f(x) = e^x \cos(x)$:

$f'(x) = (e^x)' \cdot \cos(x) + e^x \cdot (\cos(x))'$

The derivative of $e^x$ is $e^x$, and the derivative of $\cos(x)$ is $-\sin(x)$, so:

$f'(x) = e^x \cdot \cos(x) + e^x \cdot (-\sin(x))$

Thus, the equation of the derivative of f(x) is:

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

b) To find the second derivative of $f(x)$, differentiate $f'(x)$:

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

Using the product rule again, we get:

$f''(x) = (e^x)' \cdot (\cos(x) - \sin(x)) + e^x \cdot ((\cos(x) - \sin(x))')$

The derivative of $e^x$ is $e^x$, and the derivative of $\cos(x) - \sin(x)$ is $-\sin(x) - \cos(x)$, so:

$f''(x) = e^x \cdot (\cos(x) - \sin(x)) + e^x \cdot (-\sin(x) - \cos(x))$

Simplifying gives:

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

c) To find the points of inflection, set the second derivative equal to zero and solve for x:

$e^x \cdot (-\sin(x) - \cos(x)) = 0$

So, $e^x$ is never zero. This means that $-\sin(x) - \cos(x) = 0$.

We can rearrange this to get:

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

Dividing both sides by $\cos(x)$ gives:

$\tan(x) = -1$

So, the point of inflection is when $x = -\frac{\pi}{4} + n\pi$, where $n$ is an integer.

d) To determine the intervals of concavity, we examine the sign of the second derivative $f''(x) = e^x \cdot (-\sin(x) - \cos(x))$:

When $-\sin(x) - \cos(x) > 0$, the function is concave down. When $-\sin(x) - \cos(x) < 0$, the function is concave up.

We see that $-\sin(x) - \cos(x) < 0$ when $\frac{3\pi}{4} + 2n\pi < x < \frac{\pi}{4} + (2n+1)\pi$, and $-\sin(x) - \cos(x) > 0$ otherwise.

e) To determine the intervals where $f(x)$ is increasing or decreasing, we examine the sign of the first derivative $f'(x) = e^x \cdot (\cos(x) - \sin(x))$:

When $\cos(x) - \sin(x) > 0$, the function is increasing. When $\cos(x) - \sin(x) < 0$, the function is decreasing.

We see that $\cos(x) - \sin(x) < 0$ when $2n\pi < x < \frac{\pi}{4} + 2n\pi$, and $\cos(x) - \sin(x) > 0$ when $\frac{\pi}{4} + 2n\pi < x < 2n\pi$.

So, the intervals of increasing/decreasing are $(2n\pi, \frac{\pi}{4} + 2n\pi)$ and $(\frac{\pi}{4} + 2n\pi, 2n\pi)$.

This completes the solution.