f = 6√x - 9x
f' = 3/√x - 9
f(1) = -3
f(9) = 63
The avg slope is 33/4
So we want to find where the tangent has slope 33/4
3√x-9 = 33/4
3√x = 69/4
√x = 23/4
x = √23 / 2
So, is 1 < √23/2 < 9?
If so, the theorem applies.
f(x) = 6sqrt{x}- 9x n the interval [1,9].
f(c)=f(9)-f(1)/9-(1)=?
Verify that the conclusion of the Mean Value Theorem holds by computing
Now find,c in (1, 9) so that f'(c) equals the answer you just found. c=?
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