1. ## Absolutes

Find the absolute maximum and minimum values of the following function on the interval [1,3]

$\displaystyle f(x) = \frac{4x^2+9}{x}$

Do i make a table of values or what do i do?
Thanks to everyone who helps

2. what do you know about derivatives? ... i.e., what they can tell you about the behavior of a function.

3. Originally Posted by qzno
Find the absolute maximum and minimum values of the following function on the interval [1,3]

$\displaystyle f(x) = \frac{4x^2+9}{x}$

Do i make a table of values or what do i do?
Thanks to everyone who helps
The absolute maximum and minimum will either be where the derivative is 0, or on the endpoints.

The endpoints are

$\displaystyle f(1) = 13$ and $\displaystyle f(3) = 15$

The derivative is

$\displaystyle f'(x) = \frac{4x^2 - 9}{x^2}$ (use the Quotient rule).

Set this equal to 0 and solve for x.

$\displaystyle 0 = \frac{4x^2 - 9}{x^2}$

$\displaystyle 0= 4x^2 - 9$

$\displaystyle 0 = (2x + 3)(2x - 3)$

$\displaystyle 2x+3 = 0$ or $\displaystyle 2x - 3= 0$

$\displaystyle x = -\frac{3}{2}$ or $\displaystyle x = \frac{3}{2}$.

Substitute these values back into f(x) to determine their corresponding y-values.

$\displaystyle f(\frac{3}{2}) = 12$, $\displaystyle f(-\frac{3}{2}) = -12$.

So the absolute maximum is 15 which occurs at x = 3, and the absolute minumum is -12 and occurs at x = $\displaystyle -\frac{3}{2}$.

4. how is $\displaystyle x = -\frac{3}{2}$ on the interval [1,3] ?

5. Originally Posted by qzno
how is $\displaystyle x = -\frac{3}{2}$ on the interval [1,3] ?
Oops - my mistake :P.

That means the minimum is actually 12 and occurs at $\displaystyle x = \frac{3}{2}$

6. ohh ok thanks !