DBL Check My Derivative PLS

**mvho** · July 8th 2009, 04:35 PM

I have a problem where I need to find where the function is increasing/decreasing, critical values, concavity etc... But I think my first derivative may be wrong.

Question: $g(x)=\frac{4x-2.2}{(4x-4.84)^2}$

Using the quotient formula, I have the following:

$g'(x)=\frac{4*(4x-4.84)^2-2(4x-4.84)*4*(4x-2.2)}{(4x-4.84)^4}$

$g'(x)=\frac{4*(-4x-.44)}{(4x-4.84)^3}$

So to find my critical points, I set my numerator to 0.

$-4x-.44=0$
$-.44=4x$
$-.11=x$

$4x-4.84=0$
$x=1.21$

I also set my numerator to 0 (something to do with vertical asymtotes but I am not sure why we do this)

Using a number line, I think the function will be:

Decreasing: (-infinity, -.11) U (1.21, infinity)
Increasing: (-.11,1.21)

Correct?

I didn't want to go any further to test for concavity just in case my 1st derivative was wrong.

TIA

**matheagle** · July 8th 2009, 11:31 PM

algebra looks, just saw a few typos.

you mean g', not g
and you wrote 4.8 instead of 4.84 on the fifth line

**mvho** · July 9th 2009, 01:15 PM

Thanks, I've made the changes but it appears my critical numbers are wrong?

Don't I just set -4x-.44=0?

**mvho** · July 12th 2009, 03:00 PM

Ok, now I am having trouble finding my 2nd derivative and testing for concavity!

This is what I have so far:

$g"(x)=\frac{(4x-4.84)^3(-16)-(-4x-.44)(4)(3(4x-4.84)^2(4))}{(4x-4.84)^3}$

This is up to where I know I have not made any mistakes yet..

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