1. ## 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.

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

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

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

Find Critical Numbers:

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

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

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

$\displaystyle x^{2} = 1$

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

$\displaystyle x = \pm \sqrt{1}$

$\displaystyle x = -1$

$\displaystyle x = 1$

Find Min Max

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

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

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

Hey Jason76.

3. ## Re: Examining a Curve

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.

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

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

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

Find Critical Numbers:

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

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

$\displaystyle -24x^{2} = - 24$
The last line here is missing an x on the right hand side.

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

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

This is zero when $\displaystyle x=0$ or $\displaystyle x=1$

4. ## Re: Examining a Curve

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

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

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

Find Critical Numbers:

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

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

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

$\displaystyle 24x = 0$

$\displaystyle x = 0$

$\displaystyle (-x + 1) = 0$

$\displaystyle -x = -1$

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

Find Min Max

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

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

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