Question

I don't understand this :(! Please help me?

Answer 1

Let's first look at the assertion of the mean value theorem:

If
`f` is a (real-valued) function continuous on the closed interval
`[a,b]` and differentiable on the open interval
`(a,b)`, then there exists some
`c` within the open interval such that
`f'(c)=(f(b)-f(a))/(b-a)`.

So the first thing you should always check is that the given function is indeed continuous/differentiable on the given interval. Here,
`√(x-1)` is defined as long as
`x-1\ge0`, or
`x\ge1`. This contains the given interval
`[1,3]`, so we're fine. The derivative is
`\frac1{2√(x-1)}`, which also exists for all
`x\in(1,3)`, so we're fine on this front as well.

Now, we want to find the precise value(s) of
`c` that are guaranteed to exists by the MVT. All we need to do is plug in what we know to construct an equation to solve for
`c`, as is shown in the given solution:

`f'(c)=(f(b)-f(a))/(b-a)\iff\frac1{2√(c-1)}=(√(3-1)-√(1-1))/(3-1)=\frac{\sqrt2}2`

`\implies√(2(c-1))=1\implies c=\frac32`