Question

The graph of the function $f\left(x\right)=x^2+\frac{1}{x}$ is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The function is defined for all $x$ except the point $x=0$ where it has a discontinuity.
$x-$intercepts:
$f\left(x\right)=0,\Rightarrow x^2+\frac{1}{x}=0,\Rightarrow \frac{x^3+1}{x}=0,\Rightarrow x=-1$


The function is neither even, nor odd.
Check for vertical asymptotes:
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(x^2+\frac{1}{x})=-\infty$
$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}(x^2+\frac{1}{x})=+\infty$
Hence, $x=0$ (the $y-$axis) is a vertical asymptote.
Now let’s evaluate the following two limits:
$\underset{x\to \pm \infty }{lim}f\left(x\right)=\underset{x\to \pm \infty }{lim}(x^2+\frac{1}{x})=+\infty$
$k=\underset{x\to \pm \infty }{lim}\frac{f\left(x\right)}{x}=\underset{x\to \pm \infty }{lim}\frac{x^2+\frac{1}{x}}{x}=\underset{x\to \pm \infty }{lim}(x+\frac{1}{x^2})=\pm \infty$

As both these limits approach the infinity, we conclude that the function has neither horizontal, nor oblique asymptotes.

Now,
$f^{\prime }\left(x\right)=2x–\frac{1}{x^2}.$

$f^{\prime }\left(x\right)=0\Rightarrow \frac{2x^3-1}{x^2}=0,\Rightarrow$ $\{\begin{array}{c}{x}^3=\frac{1}{2}\\ x^2\ne 0\end{array}$,
$\Rightarrow$ $\{\begin{array}{c}x=\frac{1}{\sqrt[3]{32}}\\ x\ne 0\end{array}$,
$\Rightarrow x=\frac{1}{\sqrt[3]{32}}\approx 0.79$


As we can see from the sign diagram, this point is a point of local minimum. Its $y-$coordinate is equal
$f\left(\frac{1}{\sqrt[3]{32}}\right)={\left(\frac{1}{\sqrt[3]{32}}\right)}^2+\frac{1}{\frac{1}{\sqrt[3]{32}}}=\frac{1}{\sqrt[3]{34}}+\sqrt[3]{32}=\frac{1+\sqrt[3]{32}\cdot \sqrt[3]{34}}{\sqrt[3]{34}}=\frac{1+\sqrt[3]{38}}{\sqrt[3]{34}}=\frac{1+2}{\sqrt[3]{34}}=\frac{3}{\sqrt[3]{34}}\approx 1.89$


Thus, the function has a local minimum at $\left(\frac{1}{\sqrt[3]{32}},\frac{3}{\sqrt[3]{34}}\right)=\left(0.79,1.89\right).$
Now we compute the second derivative:
$f^{\prime \prime }\left(x\right)={\left(2x–\frac{1}{x^2}\right)}^{\prime }=2–\left(–2\right)x^{–3}=2+\frac{2}{x^3}=\frac{2x^3+2}{x^3}=\frac{2\left(x^3+1\right)}{x^3}.$

Determine the points where the $2nd$ derivative is zero:

$f^{\prime \prime }\left(x\right)=0,\Rightarrow \frac{2(x^3+1)}{x^3}=0,\Rightarrow$$\{\begin{array}{c}2(x^3+1)=0\\ x^3\ne 0\end{array}$,
$\Rightarrow$$\{\begin{array}{c}{x}^3=-1\\ x\ne 0\end{array}$, $\Rightarrow x=-1.$


Using the sign chart (see above) we find that $x=-1$ is a point of inflection. This point coincides with the root of the function.
Also, $f\left(x\right)$ is concave up in $x\in (-\infty ,-1)\cup (0,\infty )$