Mean Value theorem question

**hazecraze** · October 25th 2009, 07:03 PM

Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

$f(x)= \frac{x}{x+2}$ , [1,4]
---

-continuous and differentiable
f(1)=1/3
f(4)=2/3

$\frac{2/3-1/3}{4-1}$

$\frac{1/3}{3}$
=1/9

$ f'=\frac{(x+2) -x}{(x+2)^2} $

$ f'=\frac{(2)}{(x+2)^2} $ =1/9

= $(x+2)^2=18$
= $(x^2+4x-14)$

Did I make a mistake or am I suppose to use the quadratic formula to get c?

**utopiaNow** · October 25th 2009, 07:18 PM

Originally Posted by hazecraze

Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

$f(x)= \frac{x}{x+2}$ , [1,4]
---

-continuous and differentiable
f(1)=1/3
f(4)=2/3

$\frac{2/3-1/3}{4-1}$

$\frac{1/3}{3}$
=1/9

$ f'=\frac{(x+2) -x}{(x+2)^2} $

$ f'=\frac{(2)}{(x+2)^2} $ =1/9

= $(x+2)^2=18$
= $(x^2+4x-14)$

Did I make a mistake or am I suppose to use the quadratic formula to get c?

Looks fine, yes use the quadratic formula to find the $c$ in the interval $a < c < b$ ie. accept the $c$ in this interval and reject the value for $c$ which is not in this interval.

**tonio** · October 25th 2009, 07:25 PM

Originally Posted by hazecraze

Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

$f(x)= \frac{x}{x+2}$ , [1,4]
---

-continuous and differentiable
f(1)=1/3
f(4)=2/3

$\frac{2/3-1/3}{4-1}$

$\frac{1/3}{3}$
=1/9

$ f'=\frac{(x+2) -x}{(x+2)^2} $

$ f'=\frac{(2)}{(x+2)^2} $ =1/9

= $(x+2)^2=18$
= $(x^2+4x-14)$

Did I make a mistake or am I suppose to use the quadratic formula to get c?

So far so good, now continue: $(x+2)^2=18 \Longrightarrow x+2=\pm \sqrt{18}=\pm 3\sqrt{2} \Longrightarrow$ $x=\left\{\begin{array}{cc}3\sqrt{2}-2\\-3\sqrt{2}-2\end{array}\right.$

Of course, you need (and can) take only one of the above values.

Tonio

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