For f(x) = e^sin(x) use your graphing calculator to find the number of zeros for f '(x) on the closed interval [0, 2π].

May someone please explain this problem to me?

Question

For f(x) = e^sin(x) use your graphing calculator to find the number of zeros for f '(x) on the closed interval [0, 2π].

May someone please explain this problem to me?

campbell_st · Answer

if you have a parabola 

$$y = x^2 - 4x$$

the minimum for the curve is x = 2

if you find the derivative   y' = 2x - 4

then the zero of the derivative is x = 2, which is the min of the parabola. 

so graph the curve in the defined domain and find the max and min values by reading them off

they will be the zeros of the derivative

campbell_st · Answer

this site will graph it for you and show the max and mins https://www.desmos.com/calculator

anonymous · Answer

$$f(x)=e^{\sin(x)}$$
First we find the derivative
$$f'(x)=e^{\sin(x)}.\cos(x)$$
Now f'(x) will be only 0 when cos(x) is 0 because e^{sin(x)} cannot be 0 
So your question is now basically reduced to finding how many values in the interval [0,2pi] will make cos(x) 0

campbell_st · Answer

well the question is only asking you to graph the function and then identify the zeros of the derivative by looking at the graph @Nishant_Garg so no need to differentiate