Calculus and graphing

[Elisejw]

Hi, I'm having trouble with this question. Can someone please help me? Thank you!

f(x) = 1 + x/x^2-1 sketch showing any stationary points, points of inflexion or asymptotes.

[jamonwindeyer]

Hi, I'm having trouble with this question. Can someone please help me? Thank you!

f(x) = 1 + x/x^2-1 sketch showing any stationary points, points of inflexion or asymptotes.

Hey! Big question, I'll skim the details and you can let me know if you need anything elaborated!

If you differentiate and put equal to zero you can get stationary points:



Problem is, this has no solutions, so this curve has no stationary points! There is, however, a point of inflexion.  If you differentiate a second time:



We find an inflexion for \(f''(x)=0\), and this is at \((0,1)\). From here, you turn to intuition. Some things to consider:

- Intercepts?
- Where are the vertical asymptotes? That denominator can't be equal to zero...
- What does the function approach as you substitute large positive values of \(x\)? Large negative values?
- What does the function do as you substitute values near the asymptotes (EG: \(x=0.99\))

Put this together to try and approximate the shape! You are aiming for this

[Elisejw]

Hi! Thank you for the help. But, I'm having some trouble doing the second differentiation. Can you please show me how you did that? Thank you!

[RuiAce]

\begin{align*}f^\prime(x) &= \frac{-(1+x^2)}{(x^2-1)^2}\\ f^{\prime\prime}(x)&=\frac{-2x(x^2-1)^2 +(1+x^2)[4x(x^2-1)]}{(x^2-1)^4}\\ &= \frac{-2x(x^2-1) + (1+x^2)(4x)}{(x^2-1)^3}\\ &= \frac{-2x^3+2x+4x+4x^3}{(x^2-1)^3}\\ &= \frac{2x^3+6x}{(x^2-1)^3}\end{align*}

[Elisejw]

Thank you!