Find the x-coordinate where the graph of the function f(x) = e(–sinx) has a slope of 0 pie pie/4 pie/2 none of these

When is cos (x) = 0?  at 90° (aka π/2)

Hello there!

We have the function f(x) = exp(-sin(x)) and we want to see in wich point the slope is 0.

This is equivalent to see when f'(x) = 0

where f'(x) is the derivative of f(x)

Then the first step is derivate the function f(x)

if we have a function of the form g(h(x)), his derivate is:

g'(h(x))*h'(x)

In our case, g(x) = exp(x) and h(x) = - sin(x)

then f'(x) = exp(-sin(x))*(-cos(x))

Now we know that exp(-sin(x)) is never equal to 0, then we need to se when the cosine is equal to zero.

cos(90°) = 0, and 90° is equivalent to pi/2

Then f'(pi/2) = exp(-sin(pi/2))*(-cos(pi/2)) = 0

this means that the function f(x) has a slope of 0 in the point x = pi/2

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.

i really dont know

step-by-step explanation: