Question

Question #82b04

Answer 1

`c = 1`

Explanation:

I'm not sure if you meant `g(x)=sqrt(x) + 2` or `g(x) = sqrt(x+2)`. Since the interval is `[0,4]` I will assume you meant the first case (due to the easy square root), and work with that.

The Mean Value Theorem (MVT) basically says if a function `f(x)` is continuous from `[a,b]`, and is differentiable on `(a,b)`, there has to be some point `c` between `a` and `b` where:

`f'(c) = (f(b)-f(a))/(b-a)`

Think of this in the following manner: Put a dot at the point `(a, f(a))`, and then another dot at the point `(b. f(b))`. Now, connect those two points with a straight line. The MVT says there must be a point somewhere along the function `f(x)` between those two points where the tangent to `f(x)` at that point `x=c` has the exact same slope.

Here's an analogy with common terms. You are throwing a frisbee through the air to a friend of yours who is standing on a hillside that's up above you. No matter which way you throw your frisbee up or down, if you throw it and your friend catches it, at some point the frisbee had to be going in the same direction as a straight line between you and your friend.

It might start out going much steeper, but eventually gravity will take over and the frisbee will begin to level out, and eventually start flying more downwards. Sometime through that path, though, the direction through the air will be parallel to that straight line.

For this problem, first find `f(0)` and `f(4)`:

`f(0) = sqrt(0) + 2 = 0 + 2 = 2`
`f(4) = sqrt(4) + 2 = 2 + 2 = 4`

Now, let's find the slope of the line between `(0,2)` and `(2,4)`:

`(f(b)-f(a))/(b-a) = (4 - 2)/(4-0) = 2/4 = 1/2`

Lastly, the MVT says there has to be some `a<=c<=b` where `f'(c)` equals this value:

`f'(x) = 1/(2sqrt(x))`

`f'(c) = 1/2`
`1/(2sqrt(c)) = 1/2`
`2sqrt(c) = 2`
`sqrt(c) = 1`
`c = 1`