# How do you find intercepts, extrema, points of inflections, asymptotes and graph f(x)=(x^2+1)/x?

Nov 15, 2017

Analyse the function:

$f \left(x\right) = \frac{{x}^{2} + 1}{x}$

A. This is a rational function so it is defined and continuous for all real numbers except the roots of the denominator, that is in $\mathbb{R} - \left\{0\right\}$
We can note that:

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

so the function is odd.

B. The numerator ${x}^{2} + 1$ is always positive and $> 1$, so the function has no roots and $f \left(x\right) > 0$ for $x > 0$ and $f \left(x\right) < 0$ for $x < 0$. This means the graph of the function has no intercepts.

C. For $x \to 0$ we have:

${\lim}_{x \to {0}^{-}} f \left(x\right) = - \infty$

${\lim}_{x \to {0}^{+}} f \left(x\right) = \infty$

so the line $y = 0$ is a vertical asymptote.

Futhermore:

${\lim}_{x \to - \infty} f \left(x\right) = - \infty$

${\lim}_{x \to + \infty} f \left(x\right) = \infty$

so the the function does not have horizontal asymptotes. However:

${\lim}_{x \to \pm \infty} f \frac{x}{x} = {\lim}_{x \to \pm \infty} \frac{{x}^{2} + 1}{x} ^ 2 = 1$

${\lim}_{x \to \pm \infty} f \left(x\right) - x = {\lim}_{x \to \pm \infty} \frac{{x}^{2} + 1}{x} - x = {\lim}_{x \to \pm \infty} \frac{{x}^{2} + 1 - {x}^{2}}{x} = {\lim}_{x \to \pm \infty} \frac{1}{x} = 0$

so the line $y = x$ is an asympote for $x \to \pm \infty$

D. Evaluate the first derivative:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} + 1}{x}\right) = \frac{d}{\mathrm{dx}} \left(x + \frac{1}{x}\right) = 1 - \frac{1}{x} ^ 2$

thus the critical points for which $\frac{\mathrm{df}}{\mathrm{dx}} = 0$ are in $x = \pm 1$ and we can see that:

$\frac{\mathrm{df}}{\mathrm{dx}} > 0$ for $\left\mid x \right\mid > 1$ and $\frac{\mathrm{df}}{\mathrm{dx}} < 0$ for $\left\mid x \right\mid < 1$

Then $f \left(x\right)$ is monotone increasing in $\left(- \infty , - 1\right)$ and $\left(1 , \infty\right)$ while it is monotone decreasing in $\left(- 1 , 0\right)$ and $\left(0 , 1\right)$. It follows that in $x = - 1$ the function has a local maximum with value $f \left(- 1\right) = - 2$ and in $x = 1$ it has a local minimum with value $f \left(1\right) = 2$.

E. Evaluate the second derivative:

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = {d}^{2} / {\mathrm{dx}}^{2} \left(\frac{{x}^{2} + 1}{x}\right) = \frac{d}{\mathrm{dx}} \left(1 - \frac{1}{x} ^ 2\right) = \frac{2}{x} ^ 3$

We have that $\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 \ne 0$ for every $x$ so the function has no inflection points. Besides:

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 < 0$ for $x < 0$ and $\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 > 0$ for $x > 0$

so the function is concave down in $\left(- \infty , 0\right)$ and concave up in $\left(0 , \infty\right)$.

graph{ (x^2+1)/x [-20, 20, -10, 10]}