# Thread: Examining a Curve - # 3

1. ## Examining a Curve - # 3

All of this is wrong, so please look over.

$\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8$

$\displaystyle f'(x) = 12x^{2} + 18x - 54$

$\displaystyle f''(x) = 24x + 18$

Find Critical Numbers:

$\displaystyle f'(x) = 12x^{2} + 18x - 54$

$\displaystyle 12x^{2} + 18x - 54 = 0$

$\displaystyle 6(2x + 3x - 9) = 0$

$\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}$

$\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}$

$\displaystyle \dfrac{-3 \pm 9}{4}$

$\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2}$

$\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3$

Critical Numbers: $\displaystyle \dfrac{3}{2}, -3$

Find Max Min

$\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8$

$\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 = -\dfrac{370}{8}$

$\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 215$

Where does the interval increase or decrease?

$\displaystyle f'(x) = 12x^{2} + 18x - 54$

For $\displaystyle x < -3$

$\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 =$

$\displaystyle f'(-4) = 192 -72 - 54 = 66$ Positive so increasing

For $\displaystyle -3 < x <\dfrac{3}{2}$

$\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 =$

$\displaystyle f'(-1) = 12 - 18 - 54 = -60$ Negative so decreasing

For $\displaystyle \dfrac{3}{4} < x$

$\displaystyle f'(1) = 12(1)^{2} + 18(1) - 54 = - 24$ Negative so decreasing

What is the infection point?

$\displaystyle f''(x) = 24x + 18$

$\displaystyle 24x + 18 = 0$

$\displaystyle 24x = -18$

$\displaystyle x = -\dfrac{4}{3}$

$\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8$

$\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{4}{3})^{3} + 9(-\dfrac{4}{3})^{2} - 54(-\dfrac{4}{3}) + 8$

$\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{64}{67}) + 9(\dfrac{16}{9}) - 54(-\dfrac{4}{3}) + 8$

$\displaystyle f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8$

2. ## Re: Examining a Curve - # 3

Originally Posted by Jason76
All of this is wrong, so please look over.
It's not all wrong at all - just a few arithmetic errors:

Originally Posted by Jason76

$\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 = -\dfrac{370}{8}$

$\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 215$
Both of these are correct as set up but you made a mistakes with the actual calculation. Try them again.

Originally Posted by Jason76
What is the infection point?

$\displaystyle f''(x) = 24x + 18$

$\displaystyle 24x + 18 = 0$

$\displaystyle 24x = -18$
Good...

Originally Posted by Jason76

$\displaystyle x = -\dfrac{4}{3}$
The fraction is upside down! Try again.

3. ## Re: Examining a Curve - # 3

$\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8$

$\displaystyle f'(x) = 12x^{2} + 18x - 54$

$\displaystyle f''(x) = 24x + 18$

Find Critical Numbers:

$\displaystyle f'(x) = 12x^{2} + 18x - 54$

$\displaystyle 12x^{2} + 18x - 54 = 0$

$\displaystyle 6(2x + 3x - 9) = 0$

$\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}$

$\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}$

$\displaystyle \dfrac{-3 \pm 9}{4}$

$\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2}$

$\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3$

Critical Numbers: $\displaystyle \dfrac{3}{2}, -3$

Find Max Min

$\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8$

$\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =43$

$\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 287$

Where does the interval increase or decrease?

$\displaystyle f'(x) = 12x^{2} + 18x - 54$

For $\displaystyle x < -3$

$\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 =$

$\displaystyle f'(-4) = 192 -72 - 54 = 66$ Positive so increasing

For $\displaystyle -3 < x <\dfrac{3}{2}$

$\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 =$

$\displaystyle f'(-1) = 12 - 18 - 54 = -60$ Negative so decreasing

For $\displaystyle x >\dfrac{3}{2}$

$\displaystyle f'(1) = 12(3)^{2} + 18(3) - 54 = 108$ Positive so increasing

What is the infection point?

$\displaystyle f''(x) = 24x + 18$

$\displaystyle 24x + 18 = 0$

$\displaystyle 24x = -18$

$\displaystyle x = -\dfrac{4}{3}$ This looks right to me cause $\displaystyle 6$ goes into $\displaystyle 24$, $\displaystyle 6$ times and $\displaystyle 6$ goes into $\displaystyle 18$, $\displaystyle 3$ times.

$\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8$

$\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{4}{3})^{3} + 9(-\dfrac{4}{3})^{2} - 54(-\dfrac{4}{3}) + 8$

$\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{64}{67}) + 9(\dfrac{16}{9}) - 54(-\dfrac{4}{3}) + 8$

$\displaystyle f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8$

4. ## Re: Examining a Curve - # 3

Still teh same errors!

Originally Posted by Jason76
Find Max Min

$\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8$

$\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =43$
I get:
$\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =13.5 + 20.25 -81+8 = -39.25$

Originally Posted by Jason76
$\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 287$
No:
$\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = -108+81+162+8=143$

Originally Posted by Jason76
What is the infection point?

$\displaystyle f''(x) = 24x + 18$

$\displaystyle 24x + 18 = 0$

$\displaystyle 24x = -18$

$\displaystyle x = -\dfrac{4}{3}$ This looks right to me cause $\displaystyle 6$ goes into $\displaystyle 24$, $\displaystyle 6$ times and $\displaystyle 6$ goes into $\displaystyle 18$, $\displaystyle 3$ times.
If $\displaystyle 24x = -18$ then divide both sides by 24: $\displaystyle x = \frac {-18}{24} = -\frac 3 4$

By the way - I suggest that to help double-check your work you should plot the function and see whether your answers make sense.