Mean value theorem example: square root function | AP Calculus AB | Khan Academy

Let ( F(x) ) be equal to the ( \sqrt{4x - 3} ), and let ( C ) be the number that satisfies the Mean Value Theorem for ( F ) on the closed interval between 1 and 3, or ( 1 \leq x \leq 3 ). What is ( C )?

So, let's just remind ourselves what it means for ( C ) to be the number that satisfies the Mean Value Theorem for ( F ). This means that over this interval, ( C ) is a point where the slope of the tangent line at ( x = C ) (which I could write as ( F'(C) )) is equal to the slope of the secant line that connects these two points.

So this is going to be equal to the slope of the secant line that connects the points ( (3, F(3)) ) and ( (1, F(1)) ). So, this is going to be ( \frac{F(3) - F(1)}{3 - 1} ).

If you wanted to think about what this means visually, it would look something like this. So if this is our x-axis and this is 1, 2, actually let me spread it out a little bit more: 1, 2, and 3. You have ( (1, F(1)) ) right over there, so that is at the point ( (1, F(1)) ). And we could evaluate that; actually, what that's ( (1, 1) ), right? So that's going to be the point ( (1, 1) ).

Then you have the point ( (3, \ldots) ). Let's see, you're going to have ( 4 \cdot 3 = 12 - 3 = 9 ), so it's going to be ( (3, 3) ). Maybe it's right over there: ( (3, 3) ). The curve might look something like this; it might look something like that.

If you think about the slope of the line that connects these two points, the line that connects those two points—all the Mean Value Theorem tells us is that there's a point between 1 and 3 where the slope of the tangent line has the exact same slope. So if I were to eyeball it, it looks like it's right around there, although we are actually going to solve for it.

So, some point where the slope of the tangent line is equal to the slope of the line that connects these two endpoints and their corresponding function values. So that is ( C ); that would be ( C ) right over there.

So really, we just have to solve this. Let's first just find out what ( F'(x) ) is, and then we could substitute ( C ) in there and evaluate this on the right-hand side.

So, I'm going to rewrite ( F(x) ). ( F(x) ) is equal to ( (4x - 3)^{1/2} ). It makes it a little bit more obvious that we can apply the power rule and the chain rule here.

So, ( F'(x) ): ( F'(x) ) is going to be the derivative of ( (4x - 3)^{1/2} ) with respect to ( (4x - 3) ). So that is going to be ( \frac{1}{2} (4x - 3)^{-1/2} \cdot \frac{d}{dx}(4x - 3) ). The derivative of ( 4x ) with respect to ( x ) is just 4, and the derivative of -3 with respect to ( x ) is going to be 0.

So the derivative of ( 4x - 3 ) with respect to ( x ) is 4, so times 4. Thus, ( F'(x) = 4 \cdot \frac{1}{2} (4x - 3)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x - 3}} ).

Now, we could rewrite this as ( F'(C) = \frac{2}{\sqrt{4C - 3}} ).

What is that going to be equal to? That is going to be equal to—let’s see—( F(3) ) we already figured out is 3, and ( F(1) ) we already figured out is 1.

So we get ( \frac{F(3) - F(1)}{3 - 1} = \frac{3 - 1}{3 - 1} = \frac{2}{2} ), which is equal to 1. So there’s some point between 1 and 3 where the derivative at that point—the slope of the tangent line—is equal to 1.

So, let's see if we can solve this thing right over here. Well, we can multiply both sides of this by ( \sqrt{4C - 3} ).

So then we are going to get ( 2 = \sqrt{4C - 3} ). All I did is multiply both sides of this by ( \sqrt{4C - 3} ) to get rid of this in the denominator.

Now, to get rid of the radical, we can square both sides. So now we can square both sides and we get ( 4 = 4C - 3 ).

Add 3 to both sides: ( 7 = 4C ). Then divide both sides by 4. I'll go right here to do it: you're going to get ( C = \frac{7}{4} ), which is equal to ( 1.75 ).

So actually, the ( C ) value is a little bit closer; I hand drew this, it's closer to about right over there on our diagram. And actually, that looks pretty good.

I just hand drew this curve, so it’s definitely not exact, but anyway, hopefully, that gives you a sense of what’s going on here.

We’re just saying, hey, the Mean Value Theorem gives us some ( C ) where the slope of the tangent line is the same as the slope of the line that connects ( F(1) ) and ( F(3) ).