# Math Help - Help with graph please- concavity, relative extrema, etc.

1. ## Help with graph please- concavity, relative extrema, etc.

2x^2+1
--------
2x^2-3x

Show any asymptotes, relative extrema, inflection points, concavity, and where the function is increasing or decreasing.

I'm really having trouble with this , any help would be GREATLY appreciated.

Thanks guy/girls.

2. Originally Posted by BigMath
2x^2+1
--------
2x^2-3x

Show any asymptotes, relative extrema, inflection points, concavity, and where the function is increasing or decreasing.

I'm really having trouble with this , any help would be GREATLY appreciated.

Thanks guy/girls.
Set $f'(x)=0$ to locate any extrema.

If $f'(x)$ is undefined at some number $x=c$, then $x=c$ is a vertical asymptote.

If $\lim_{x\to\pm\infty}f(x)=L$, then $f(x)=L$ is a horizontal asymptote.

Get started with that and then show me what you've got.

3. That's basically all I know how to do:

For H.A. I got: 1

For V.A. I got: 3/2, 0

For the increasing/decreasing do i solve for first derivative than make it equal 0? Or do I use the V.A. points and plug in to the first derivative.

4. Originally Posted by BigMath

For the increasing/decreasing do i solve for first derivative than make it equal 0?
$f'(x)>0$ means that $f$ is increasing.

$f'(x)<0$ means that $f$ is decreasing.

So, you must solve these inequalities to determine the invervals over which $f$ is increasing or decreasing.

5. My teacher wanted us to make a chart, with infinity to certain number, certain number to another, and that number to negative infinity and to plug in numbers to get increasing/decreasing. Can you show me?

6. Originally Posted by BigMath
That's basically all I know how to do:

For H.A. I got: 1 correct

For V.A. I got: 3/2, 0 correct

For the increasing/decreasing do i solve for first derivative than make it equal 0? Or do I use the V.A. points and plug in to the first derivative.
.

7. For the numbers you plug in the first derivative, to see if they are greater or less than 0, where do you get those numbers from? Do you use the V.A.?

8. Are they

-infinity, -1.115

-1.115, 0.448

0.448, 0

0, 3/2

3/2, infinity

Than plug in numbers in between to see increasing or decreasing?

9. Originally Posted by BigMath
My teacher wanted us to make a chart, with infinity to certain number, certain number to another, and that number to negative infinity and to plug in numbers to get increasing/decreasing. Can you show me?
I'm not skilled enough with LaTeX to make a chart on this forum, but the idea is to choose test values between the critical numbers to see if $f'>0$ or $f'<0$.

$f(x)=\frac{2x^2+1}{2x^3-3x}$

$f'(x)=\frac{(2x^2-3x)(4x)-(2x^2+1)(6x^2-3)}{(2x^3-3x)^2}$

$=\frac{(8x^3-12x^2)-(12x^4-3)}{(2x^3-3x)^2}$

$=\frac{(8x^3-12x^2)-(12x^4-3)}{(2x^3-3x)^2}=0$ implies

$(8x^3-12x^2)-(12x^4-3)=0$

$8x^3-12x^2-12x^4+3=0$

$12x^4-8x^3+12x^2-3=0$ Which is really ugly. Are you sure that this is the right problem?

10. The bottom is 2x^2-3x

Derivative comes out to like:

-6x^2-4x+3
----------
(2x^2-3x)^2

No idea why that is underlined

So when I solve for 0, I get

-6x^2-4x+3=0

I used the quadratic formula.

11. Originally Posted by BigMath

The bottom is 2x^2-3x

Derivative comes out to like:

-6x^2-4x+3
----------
(2x^2-3x)^2

No idea why that is underlined

So when I solve for 0, I get

-6x^2-4x+3=0

I used the quadratic formula.

ooops. My bad. OK, yeah, so it doesn't factor. Using the quadratic formula should give the two answers that you've given before.

Yeah. You're on the right track. So, make a table that describes the sign of f prime at some test value on the intervals that you have already discovered a couple of posts ago before I went in left field with doing the wrong problem.

12. Okay I got this:

Than I plug it into the first derivative to see increasing or decreasing:

f1(x) f(x)
-infinity, -1.115 - dec

-1.115, 0.448 + inc

0.448, 0 + inc

0, 3/2 - dec

3/2, infinity - dec

Are those the right points I would use? So can you help me check this so far:

Increasing/decreasing: see above

Relative max: (0, f(0))

Relative min: (-1.115, f(-1.115))

Right or wrong?