# Examining a Curve

• October 29th 2013, 01:27 AM
Jason76
Examining a Curve
All of this is wrong. Here is the actual question: "Find the local maximum and minimum values of f using both the First and Second Derivative Tests." Please look thru this to see what is mistaken.

$f(x) = -8x^{3} + 12x^{2} + 8$

$f'(x) = -24x^{2} + 24x$

$f''(x) = -48x + 24$

Find Critical Numbers:

$f'(x) = -24x^{2} + 24x$

$-24x^{2} + 24x = 0$

$-24x^{2} = - 24$

$x^{2} = 1$

$\sqrt{x^{2}} = \sqrt{1}$

$x = \pm \sqrt{1}$

$x = -1$

$x = 1$

Find Min Max

$f(x) = -8x^{3} + 12x^{2} + 8$

$f(-1) = -8(-1)^{3} + 12(-1)^{2} + 8 = 28$

$f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12$
• October 29th 2013, 02:57 AM
chiro
Re: Examining a Curve
Hey Jason76.

• October 29th 2013, 12:03 PM
TwoPlusTwo
Re: Examining a Curve
Quote:

Originally Posted by Jason76
All of this is wrong. Here is the actual question: "Find the local maximum and minimum values of f using both the First and Second Derivative Tests." Please look thru this to see what is mistaken.

$f(x) = -8x^{3} + 12x^{2} + 8$

$f'(x) = -24x^{2} + 24x$

$f''(x) = -48x + 24$

Find Critical Numbers:

$f'(x) = -24x^{2} + 24x$

$-24x^{2} + 24x = 0$

$-24x^{2} = - 24$

The last line here is missing an x on the right hand side.

You can also write $f'(x)$ as:

$f'(x) = 24x(1-x)$

This is zero when $x=0$ or $x=1$
• October 29th 2013, 01:12 PM
Jason76
Re: Examining a Curve
$f(x) = -8x^{3} + 12x^{2} + 8$

$f'(x) = -24x^{2} + 24x$

$f''(x) = -48x + 24$

Find Critical Numbers:

$f'(x) = -24x^{2} + 24x$

$-24x^{2} + 24x = 0$

$24x(-x + 1) = 0$ a different approach

$24x = 0$

$x = 0$

$(-x + 1) = 0$

$-x = -1$

$x = 1$ and $x = 0$ (Critical Number)

Find Min Max

$f(x) = -8x^{3} + 12x^{2} + 8$

$f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12$ - Max

$f(0) = -8(0)^{3} + 12(0)^{2} + 8 = 8$ - Min