# Thread: Quotient Rule - overall max value of function

1. ## Quotient Rule - overall max value of function

Hi,

I'm stuck on this and would be grateful for any help.

I need to find the overall maximum value (correct to two decimal placeds) of the function:

f(x) = 20x / 5 + x^2

on the interval (0, 6)

2. Hello buddy.

Originally Posted by looking0glass
Hi,

I'm stuck on this and would be grateful for any help.

I need to find the overall maximum value (correct to two decimal placeds) of the function:

f(x) = 20x / 5 + x^2

on the interval (0, 6)

You do! You already know you must use the quotient rule. Why is that? Because if you want to find the critical points of a function f(x), you need to consider the derivative : f'(x)

Finding f'(x) - use quotient rule

$f'(x) = \frac{20*(5+x^2)-2x*20x}{(5+x)^2} = \frac{20(5 - x^2)}{(x^2 + 5)^2}$

To find the critical points solve

f'(x) = 0. Thus

$\frac{20(5 - x^2)}{(x^2 + 5)^2} = 0$

multiply by (x^2+5)²

$20(5-x^2) = 0$

<=> $5-x^2 = 0$

and then you find $x_E$.

To show that it is a max, solve $f''(x_E)$; $f''(x_E)$ must be < 0

To find the y-coordinate of your critical point, you need to solve $f(x_E)$

Yours
Rapha

3. Hi thanks!

Yep! The critical point is $(\sqrt{5} \ , \ 4.47)$, so you are correct..