what are the conditions that the function must satisfy to apply the MVT?
Does √x meet the conditions on the given interval?
Yes, since √x is continuous on [0,4] and differentiable on (0,4).
(Why is it not differentiable on the closed interval?)
So, the MVT says that there will be some point x=c in (0,4) such that the slope of the line joining the two points on the graph at the ends of the interval is the same as the slope of the tangent line at x=c. That is,
(f(4)-f(0))/(4-0) = f'(c)
That is,
f'(c) = 2/4 = 1/2
So, where is f' = 1/2 ?
f'(x) = 1/(2√x)
so we need 1/(2√c) = 1/2
c=1
Hmmm. Not one of the choices. But, we see from the graphs that
the tangent line at (1,1) does indeed have slope 1/2
https://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9Ax,+y%3D1%2F2+(x-1)%2B1,y%3Dx%2F2+for+0%3C%3Dx%3C%3D4
Determine if the Mean Value Theorem for Integrals applies to the function f(x) = √x on the interval [0, 4]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
a) No, the theorem does not apply
b) Yes, x=16/9
c) Yes, x=1/4
d) Yes, x=1/16
2 answers
x = 16/9
8/3 = 2sqrt(x)
4/3 = sqrt(x)
16/9 = x
8/3 = 2sqrt(x)
4/3 = sqrt(x)
16/9 = x