Asked by Kasie
Using the mean value theorem;
F'(x) = f(b)-f(a) / b-a
f(x)=x^2-8x+3; interval [-1,6]
F'(x) = f(b)-f(a) / b-a
f(x)=x^2-8x+3; interval [-1,6]
Answers
Answered by
Steve
I assume you want to find c such that f'(c) = (f(6)-f(-1))/7
nothing simpler:
f'(x) = 2x-8
f(6) = -9
f(-1) = 12
so, we want f'(c) = -21/7 = -3
2x-8 = -3
x = 5/2
nothing simpler:
f'(x) = 2x-8
f(6) = -9
f(-1) = 12
so, we want f'(c) = -21/7 = -3
2x-8 = -3
x = 5/2
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