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

Results 1 to 6 of 6

- Nov 11th 2008, 12:41 PM #1

- Joined
- Oct 2008
- Posts
- 109

- Nov 11th 2008, 01:53 PM #2

- Nov 11th 2008, 01:58 PM #3
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 = $\displaystyle -\frac{3}{2}$.

- Nov 11th 2008, 02:02 PM #4

- Joined
- Oct 2008
- Posts
- 109

- Nov 11th 2008, 02:06 PM #5

- Nov 11th 2008, 02:11 PM #6