Question

Is `f(x)=e^-xcos(-x)` increasing or decreasing at `x=pi/6`?

Answer 1

First, note that because the cosine function is and even function , then `cos(-x) = cos(x)`.

Explanation:

The function will be increasing if the first derivative is positive at `x=pi/6`. It will be decreasing, if it is negative.

Also note that `e^(-x)` is ALWAYS positive regardless what value of x you insert.

Let's find `f'` using the product rule ...

`f'=(-e^(-x))cosx -(e^-x)sinx`

Both sine and cosine of `pi/6` or `30^o` is a positive number. And, as stated above, `e^-x` is always positive.

So, `f'=`(negative)(positive)-(positive)(positive) = ( negative )

Answer: f(x) is decreasing at `pi/6`

hope that helped