Verify the conditions for Rolle's Theorem for the function f(x)=x^2/(8x-15) on the interval [3,5] and find c in this interval such that f'(c)=0
I verified that f(a)=f(b) and calculated f'(x)= (8x^2 -30x)/64x^2 -240x +225)
But I'm having trouble finding c when that derivative is equal to 0.
I find it a bit easier not to expand the derivative
f'(x) = 2x(4x-15)/(8x-15)^2
Clearly f'=0 when x is 0 or 15/4
So, f'(15/4)=0, and 3 < 15/4 < 5
f'(0)=0 also, but 0 is not in [3,5]
The graph at
http://www.wolframalpha.com/input/?i=x^2%2F%288x-15%29+for+3+%3C%3D+x+%3C%3D+5
clearly shows that f'(3.75) is zero.
To find the value of c in the interval [3,5] such that f'(c) = 0, we need to solve the equation f'(x) = (8x^2 - 30x)/(64x^2 - 240x + 225) = 0.
First, let's simplify the equation:
8x^2 - 30x = 0
Factor out the common factor of 2x:
2x(4x - 15) = 0
Now, set each factor equal to zero and solve for x:
2x = 0 or 4x - 15 = 0
For the first factor:
2x = 0
x = 0
For the second factor:
4x - 15 = 0
4x = 15
x = 15/4 or 3.75
Now, we have two potential values of c: 0 and 3.75. However, we need to consider if these values of c belong to the interval [3,5].
Since c = 0 is not in the interval [3,5], we can disregard it.
Therefore, the value of c in the interval [3,5] such that f'(c) = 0 is c = 3.75.