# Thread: Mean Value theorem question

1. ## Mean Value theorem question

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?

2. 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.

3. 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