Consider the function http://www.imathas.com/cgi-bin/mimet...D%5Cleq%7B5%7D.

This function has an absolute minimum value equal to

and an absolute maximum value equal to

Printable View

- Nov 28th 2012, 11:25 AMapple111derivative function
Consider the function http://www.imathas.com/cgi-bin/mimet...D%5Cleq%7B5%7D.

This function has an absolute minimum value equal to

and an absolute maximum value equal to - Nov 28th 2012, 12:34 PMskeeterRe: derivative function
what have you done with this problem?

- Nov 29th 2012, 05:44 PMx3bnmRe: derivative function
Absolute max or min is where first derivative is zero.

$\displaystyle f(x) =& 2x^3 + 9x^2 -168x + 3$

$\displaystyle \begin{align*} \frac{df}{dx} =& 6x^2 + 18x -168 = 0 \\ &x^2 + 3 x - 28 = 0.......\text{[after dividing both sides by } 6 \text{ ]}\\ &(x + 7)(x - 4) = 0......\text{[after factorization]}\end{align*}$

So absolute max or min can be at $\displaystyle x = -7$ or $\displaystyle x = 4$. But to find out which one which you plug-in the values into the function and find out which one's larger than the other.

Now plug-in $\displaystyle x = -7$ and separately $\displaystyle x = 4$ into $\displaystyle f(x)$ you'll find that $\displaystyle f(-7) > f(4)$

The function has absolute max at $\displaystyle x = -7$

And has absolute min at $\displaystyle x = 4$

You can check it in the following graph to be sure:

Attachment 25984 - Nov 29th 2012, 07:56 PMProve ItRe: derivative function
- Nov 29th 2012, 09:06 PMx3bnmRe: derivative function