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.
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$\displaystyle f'(x)=0$ to locate any extrema.
If $\displaystyle f'(x)$ is undefined at some number $\displaystyle x=c$, then $\displaystyle x=c$ is a vertical asymptote.
If $\displaystyle \lim_{x\to\pm\infty}f(x)=L$, then $\displaystyle f(x)=L$ is a horizontal asymptote.
Get started with that and then show me what you've got.
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 $\displaystyle f'>0$ or $\displaystyle f'<0$.
$\displaystyle f(x)=\frac{2x^2+1}{2x^3-3x}$
$\displaystyle f'(x)=\frac{(2x^2-3x)(4x)-(2x^2+1)(6x^2-3)}{(2x^3-3x)^2}$
$\displaystyle =\frac{(8x^3-12x^2)-(12x^4-3)}{(2x^3-3x)^2}$
$\displaystyle =\frac{(8x^3-12x^2)-(12x^4-3)}{(2x^3-3x)^2}=0$ implies
$\displaystyle (8x^3-12x^2)-(12x^4-3)=0$
$\displaystyle 8x^3-12x^2-12x^4+3=0$
$\displaystyle 12x^4-8x^3+12x^2-3=0$ Which is really ugly. Are you sure that this is the right problem?
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.
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?