1. ## Mean Value Theorem

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?

2. 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 $\displaystyle f(2)=8$ and $\displaystyle f(0)=0$.

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

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

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

So what is $\displaystyle c$?

3. Originally Posted by Chris L T521
Note $\displaystyle f(2)=8$ and $\displaystyle f(0)=0$.

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

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

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

So what is $\displaystyle c$?
C=1

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