Is there a way to do this problem without a calculator

**zhangxupage2** · January 17th 2011, 03:07 PM

Determine whether the Mean Value Theorem applies to f(x) = 3x – x^2 on the interval [2, 3]. If it can be applied, find all value(s) of c in the interval that satisfy the Mean Value Theorem.

This is what I did

$\int f(x) dx$

$3x^2/2 - x^3/3$

$F(3)=3*2^2/2-2^3/3=4.5$
$F(2)=3*2^2/2-2^3/3=3+1/3$
$4.5-(3+1/3)=7/6$

$f(c)=3c-c^2$

I used my graphing calculator and found the intersection between f(c) and 7/6, and the answer is 2.5408.

Did I do it correctly? If I did, is there a way to do it without a calculator?

**skeeter** · January 17th 2011, 04:15 PM

Originally Posted by zhangxupage2

Determine whether the Mean Value Theorem applies to f(x) = 3x – x^2 on the interval [2, 3]. If it can be applied, find all value(s) of c in the interval that satisfy the Mean Value Theorem.

This is what I did

$\int f(x) dx$

$3x^2/2 - x^3/3$

$F(3)=3*2^2/2-2^3/3=4.5$
$F(2)=3*2^2/2-2^3/3=3+1/3$
$4.5-(3+1/3)=7/6$

$f(c)=3c-c^2$

I used my graphing calculator and found the intersection between f(c) and 7/6, and the answer is 2.5408.

Did I do it correctly? If I did, is there a way to do it without a calculator?

MVT for integrals, yes ...

$\displaystyle 3c - c^2 = \frac{7}{6}$

$\displaystyle 0 = c^2 - 3c + \frac{7}{6}$

quadratic formula ...

$\displaystyle x = \frac{3 + \sqrt{9 - \frac{14}{3}}}{2}$

**zhangxupage2** · January 17th 2011, 04:41 PM

Thanks a lot!

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