# Thread: Graphing with Calculus and Calculators

1. ## Graphing with Calculus and Calculators

I am very annoyed with these certain problems because I have spend 2.5hrs on them and do not have the full answer... If someone can show me their thinking process for these I would very much appreciate it...

For the first two i basically have the answer, but not the asymptope parts..

Produce Graphs for f that reveal all the important aspects of the curve. In particular you should use graphs of f' and f'' to estimate the intervals of increase and decrease, extreme intervals, intervals of concavity, and inflection pts...
1) f(x) = (X^2-3X-5)^(1/3)
2) f(x) = X/(X^3-X^2-4X+1)

and for the third, "Produce graphs of f that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, use calculus to find these intervals exactly...
3) f(x) = 1+ 1/x + 8/x^2 + 1/X^3

If you could work out these three problems, I would be very thankful. I have graphs already, and i understand dirivitives etc... However for the first two the asymptopes are giving me problems and for the 3rd my exact answer is coming out incorrect each time.. Thank you for your time... any response would be very helpful.

2. If you don't know how to draw the graph using f'(x) and f''(x), take a look at these videos. They explain it quite well. If you know how to find asymptotes, maximas, minimas and other important points, you can watch only the 17th vid. Otherwise, watch the topic you need help with.

Calculus Videos

3. For the third problem, can someone please work out how to get the exact numbers? my calculations keep coming out as being wrong.
Thanks!

4. Here's a few hints.

If we mulltiply by x^3, we get $x^{3}+x^{2}+8x+1$

Set to 0 and solve for x, we find where it crosses the x-axis at

$x = -0.126753752867.......$

$f'(x)=\frac{-3}{x^{4}}-\frac{16}{x^{3}}-\frac{1}{x^{2}}=\frac{-(x^{2}+16x+3)}{x^{4}}$

Setting to 0 and solving for x to find the extrema, we get:

$x = -(\sqrt{61}+8), \;\ x = \sqrt{61}-8$

For the second derivative:

$f''(x)=\frac{12}{x^{5}}+\frac{48}{x^{4}}+\frac{2}{ x^{3}}=\frac{2(x^{2}+24x+6)}{x^{5}}$

Set to 0 and solve for x, gets us: $x = -(\sqrt{138}+12), \;\ x = \sqrt{138}-12$

Find your horizontal asymptotes, let $x\rightarrow{\infty}$

For any vertical asymptotes, what does the limit approach as x approaches 0 from the right or left?.

Always make a graph. That helps.

That should be a good start.