f(5) = 7 + 8 sqrt 5
f(1) = 7 + 8 sqrt 1
f(5) - f(1) = 8(sqrt 5 - 1) = 9.89
9.89/4 = mean slope = 2.47
so where does f'(x) = 2.47 ????
f'(x) = 4 /x^(1/2)
so
2.47 = 4/x^.5
x^.5 = 4/2.47 = 1.62
x = 2.62
check that please.
Consider the function f(x)=8(x^(1/2))+7 on the interval [1,5].
By the Mean Value Theorem, we know there exists a c in the open interval (1,5) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
1 answer
1 answer