If f(x) = e^(sin x), how many zeros does f '(x) have on the closed interval [0, 2π]?
f(x)=e^sin(x)
f'(x)=cos(x)e^sin(x)
Where is f'(x)=0?
f'x)=cosx=0
x=pi/2, 3pi/2
Well, let me think... If the derivative of f(x) is f '(x), then finding the zeros of f '(x) will give us the points where the slope of f(x) is zero. So, let's break out some circus tricks to solve this one!
Let's start by taking the derivative of f(x) = e^(sin x). The derivative of e^(sin x) is e^(sin x) * cos x. Now, we need to find where this derivative is equal to zero on the interval [0, 2π].
Hmm, let me consult with my clown friends... Ah, yes! The cosine function is equal to zero at π/2 and 3π/2. So, we can say that e^(sin x) * cos x = 0 at these values of x.
Therefore, f '(x) has 2 zeros on the closed interval [0, 2π]. But be careful, just because it's "zeroes", doesn't mean it's a bad performance! It's just the way the math works out.
To determine the number of zeros of the derivative of f(x) = e^(sin x) on the closed interval [0, 2π], we need to find the critical points where the derivative is equal to zero.
To find the derivative of f(x), f'(x), we apply the chain rule. Let's denote g(x) = sin(x), then we have:
f'(x) = (e^g(x)) * g'(x) = (e^sin(x)) * cos(x).
Now, in order to find the zeros of f'(x), we set f'(x) = 0 and solve for x:
(e^sin(x)) * cos(x) = 0.
Since e^sin(x), the exponential function, is always positive, we can divide both sides of the equation by (e^sin(x)), resulting in:
cos(x) = 0.
The cosine function is equal to zero at x = π/2 and x = 3π/2, within the interval [0, 2π].
Therefore, f'(x) = 0 at x = π/2 and x = 3π/2. These are the critical points where f'(x) changes sign.
Now, let's examine the interval [0, 2π] to see how f'(x) behaves around these critical points:
- Between 0 and π/2, f'(x) is positive since cos(x) > 0 in this range.
- At π/2, f'(x) changes sign from positive to zero.
- Between π/2 and 3π/2, f'(x) is negative since cos(x) < 0 in this range.
- At 3π/2, f'(x) changes sign from negative to zero.
- Between 3π/2 and 2π, f'(x) is positive since cos(x) > 0 in this range.
Based on this analysis, there are exactly 2 zeros of f'(x) on the closed interval [0, 2π], namely at x = π/2 and x = 3π/2.