Mean Value Theorem

**KarlosK** · May 17th 2010, 07:18 AM

Apply the mean value theorem to the function f(x)= x^2 +2x on the interval [0,2] to determine any valid conclusion the theorem would make.

I know the formula is f(b)-f(a) / b-a = f'(c)

I ended up with (4+2)-(2) / (2-0) = 2

So do I take that 2 and plug it into the derivative of the equation. I did that and got 2(2)+2=6

I am not sure if I did that right, or what it says about the equation?

**Chris L T521** · May 17th 2010, 07:22 AM

Originally Posted by KarlosK

Apply the mean value theorem to the function f(x)= x^2 +2x on the interval [0,2] to determine any valid conclusion the theorem would make.

I know the formula is f(b)-f(a) / b-a = f'(c)

I ended up with (4+2)-(2) / (2-0) = 2

So do I take that 2 and plug it into the derivative of the equation. I did that and got 2(2)+2=6

I am not sure if I did that right, or what it says about the equation?

Note $f(2)=8$ and $f(0)=0$ .

Now, $\frac{f(2)-f(0)}{2-0}=\frac{8-0}{2-0}=4$ .

Also note that $f^{\prime}(x)=2x+2\implies f^{\prime}(c)=2c+2$

Therefore, by MVT, $\exists\,c\in[0,2]: 2c+2=4$ .

So what is $c$ ?

**KarlosK** · May 17th 2010, 07:35 AM

Originally Posted by Chris L T521

Note $f(2)=8$ and $f(0)=0$ .

Now, $\frac{f(2)-f(0)}{2-0}=\frac{8-0}{2-0}=4$ .

Also note that $f^{\prime}(x)=2x+2\implies f^{\prime}(c)=2c+2$

Therefore, by MVT, $\exists\,c\in[0,2]: 2c+2=4$ .

So what is $c$ ?

C=1

So you plug the first numbers into the original function... and only plug the value in the first equation into the derivative.

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