# Graphing a rational function

• Mar 31st 2010, 12:36 PM
Jools
Graphing a rational function
Got another problem...

Given f(x) = (x+5)/(2x-4)

Calculate the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points, and concavity.

I've calulated the first and second derivatives and come up with this:

f '(x) = (2x^2 + 4x - 30) / (2x-4)^2

f ''(x) = 8(x-17) / (2x-4)^3

I'm pretty sure those are right... but maybe not. In orer to find the intervals of increase and decrease I should find where f '(x) = 0 right? If that's the case I may need a little push to get that figured out. Based on the second derivative, the inflection point should be x=17 right? That doesn't seem right either... I'm all screwed up on this one I think...
• Mar 31st 2010, 02:41 PM
skeeter
Quote:

Originally Posted by Jools
Got another problem...

Given f(x) = (x+5)/(2x-4)

Calculate the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points, and concavity.

I've calulated the first and second derivatives and come up with this:

f '(x) = (2x^2 + 4x - 30) / (2x-4)^2

f ''(x) = 8(x-17) / (2x-4)^3

I'm pretty sure those are right... but maybe not. In orer to find the intervals of increase and decrease I should find where f '(x) = 0 right? If that's the case I may need a little push to get that figured out. Based on the second derivative, the inflection point should be x=17 right? That doesn't seem right either... I'm all screwed up on this one I think...

$\displaystyle f'(x)$ is incorrect.
• Mar 31st 2010, 04:19 PM
Jools
Quote:

Originally Posted by skeeter
$\displaystyle f'(x)$ is incorrect.

Gotcha... I just checed f '(x) again and see where I made the errror. Does this look better for f '(x)?

-14 / (2x-4)^2

Thanks!
• Mar 31st 2010, 04:27 PM
skeeter
Quote:

Originally Posted by Jools
Gotcha... I just checed f '(x) again and see where I made the errror. Does this look better for f '(x)?

28 / (2x-4)^3

Thanks!

$\displaystyle f(x) = \frac{x+5}{2x-4}$

$\displaystyle f(x) = \frac{1}{2} \cdot \frac{x+5}{x-2}$

$\displaystyle f'(x) = \frac{1}{2} \cdot \frac{(x-2)(1) - (x+5)(1)}{(x-2)^2}$

$\displaystyle f'(x) = \frac{1}{2} \cdot \frac{x-2 - x - 5}{(x-2)^2}$

$\displaystyle f'(x) = -\frac{7}{2(x-2)^2}$

$\displaystyle f'(x) = -\frac{7}{2}(x-2)^{-2}$

$\displaystyle f''(x) = 7(x-2)^{-3} = \frac{7}{(x-2)^3}$
• Mar 31st 2010, 04:38 PM
Jools
Quote:

Originally Posted by skeeter
$\displaystyle f(x) = \frac{x+5}{2x-4}$

$\displaystyle f(x) = \frac{1}{2} \cdot \frac{x+5}{x-2}$

$\displaystyle f'(x) = \frac{1}{2} \cdot \frac{(x-2)(1) - (x+5)(1)}{(x-2)^2}$

$\displaystyle f'(x) = \frac{1}{2} \cdot \frac{x-2 - x - 5}{(x-2)^2}$

$\displaystyle f'(x) = -\frac{7}{2(x-2)^2}$

$\displaystyle f'(x) = -\frac{7}{2}(x-2)^{-2}$

$\displaystyle f''(x) = 7(x-2)^{-3} = \frac{7}{(x-2)^3}$

Here's how I came up with f '(x)... Can you show me where I went wrong?

f(x) = (x+5) / (2x-4)

f(x) = (x+5)(2x-4)^-1
f '(x) = (x+5)' (2x-4)^-1 + (x+5)(2x-4)^-1'
= (1)(2x-4)^-1 + (x+5)(-1)(2x-4)^-2(2)
= (2x-4)^-1 - 2(x+5)(2x-4)^-2
= 1 / (2x-4) - 2(x+5) / (2x-4)^2
= (2x-4)/(2x-4)^2 - 2(x+5)/2x-4)^2
= 2x-4 - 2(x+5) / (2x-4)^2
= 2x-4-2x-10 / (2x-4)^2
f '(x) = -14 / (2x-4)^2

Did I go wrong somewhere? It's close to what you have... Only I've gotten -14 as the numerator instead of -7.
• Mar 31st 2010, 05:10 PM
skeeter
it's the same thing, I just simplified a bit more ...

$\displaystyle \frac{-14}{(2x-4)^2} = \frac{-14}{[2(x-2)]^2} = \frac{-14}{4(x-2)^2}= \frac{-7}{2(x-2)^2}$
• Mar 31st 2010, 05:21 PM
Jools
Quote:

Originally Posted by skeeter
it's the same thing, I just simplified a bit more ...

$\displaystyle \frac{-14}{(2x-4)^2} = \frac{-14}{[2(x-2)]^2} = \frac{-14}{4(x-2)^2}= \frac{-7}{2(x-2)^2}$

I don't have the 4 in the denominator though...
• Mar 31st 2010, 05:38 PM
skeeter
Quote:

Originally Posted by Jools
I don't have the 4 in the denominator though...

and you shouldn't ... look at the last equation again.
• Mar 31st 2010, 06:10 PM
Jools
Quote:

Originally Posted by skeeter
and you shouldn't ... look at the last equation again.

Sorry I'm missing something here...

I'm getting

f '(x) = -14 / (2x-4)^2

You have -14 / [2(x-2)]^2 which is working out to -14 / 4(x-2)^2. Should you not multiply out the 2(x-2) portion first before squaring it?
• Apr 1st 2010, 04:36 AM
skeeter
Quote:

Originally Posted by Jools
Sorry I'm missing something here...

I'm getting

f '(x) = -14 / (2x-4)^2

You have -14 / [2(x-2)]^2 which is working out to -14 / 4(x-2)^2. Should you not multiply out the 2(x-2) portion first before squaring it?

working with the denominator ...

$\displaystyle (2x-4)^2 = [2(x-2)]^2 = 2^2(x-2)^2 = 4(x-2)^2$

now put that under -14 ...

$\displaystyle \frac{-14}{4(x-2)^2} = \frac{-7}{2(x-2)^2} = \frac{-14}{(2x-4)^2}$