The function defined below satisfies Rolle's Theorem on the given interval. Find the value of c in the interval (-3, -2) where f'(c)=0.
f(x) = x + 6x-1 , [-3, -2]
Round your answer to two decimal places.
I dislike f(x). It is linear, so cannot satisfy the theorem.
maybe x^2 + 6 x - 1 ?
I thought of that, but in that case, f(-3) ≠ f(-2), so the theorem does not apply.
i think its x+6x^-1
To find the value of c in the interval (-3, -2) where f'(c) = 0, we need to find the derivative of f(x) first.
The function f(x) is given as:
f(x) = x + 6x-1
To find the derivative, we differentiate each term separately using the power rule and the constant rule:
f'(x) = 1 + 6(-1)x^(1-1) = 1 - 6x^0 = 1 - 6 = -5
Now that we have the derivative, we can set it equal to 0 and solve for c:
-5 = 0
Since -5 is never equal to 0, there is no value of c in the interval (-3, -2) where f'(c) = 0. Therefore, there is no solution for this problem.
I hope this explanation helps! Let me know if you have any further questions.