# How do you graph f(x)=1/x-3x^3 using the information given by the first derivative?

Jun 10, 2018

Look for zeros on the first and second derivatives.

#### Explanation:

Start by finding the first and second derivative.

$f \left(x\right) = \frac{1}{x} - 3 {x}^{3} = {x}^{- 1} - 3 {x}^{3}$

rewrite $f \left(x\right)$ such that it fits the power rule.

$f ' \left(x\right) = - {x}^{- 2} - 9 {x}^{2} = - \frac{1}{{x}^{2}} - 9 {x}^{2}$

Now look at the first derivative: both $- \frac{1}{x} ^ 2$ and $- 9 {x}^{2}$ are negative for real values of x. Additionally, the first derivative is not defined for $x = 0$ and demonstrates asymptotic behavior for that value. Thus $f \left(x\right)$ shall consist of two branches and given that $f ' \left(x\right) < 0$, $f \left(x\right)$ is decreasing over both branches.

Zeros and signs on the second derivative give additional information about the concavity of and inflection points (if any) on $f \left(x\right)$.

$f ' ' \left(x\right) = 2 {x}^{- 3} - 18 x = \frac{2}{{x}^{3}} - 18 x$

Solving $f ' ' \left(x\right) = 0$ gives

${x}^{4} = \frac{1}{9}$,
$x = \pm \frac{1}{\sqrt{3}}$

$f ' ' \left(x\right)$ shows asymptotic behavior at $x = 0$ so checking the value of $f ' ' \left(x\right)$ on the four open intervals

• $\left(- \infty , - \frac{1}{\sqrt{3}}\right)$
• $\left(- \frac{1}{\sqrt{3}} , 0\right)$
• $\left(0 , \frac{1}{\sqrt{3}}\right)$, and
• $\left(\frac{1}{\sqrt{3}} , \infty\right)$

shall give a comprehensive conclusion on the sign of $f ' ' \left(x\right)$ over the whole real domain.

• $f ' ' \left(x\right) > 0$ for $x \in \left(- \infty , - \frac{1}{\sqrt{3}}\right)$ and $x \in \left(0 , \frac{1}{\sqrt{3}}\right)$;
• $f ' ' \left(x\right) < 0$ for $x \in \left(- \frac{1}{\sqrt{3}} , 0\right)$ and $x \in \left(\frac{1}{\sqrt{3}} , 0\right)$.

Conclusion:

• $f \left(x\right)$ demonstrates asymptotic behavior on $x = 0$
• $f \left(x\right)$ is decreasing on both of its branches (from 1st derivative).

taking information from the second derivative into account:

• $f \left(x\right)$ is concave upward on $x \in \left(- \infty , - \frac{1}{\sqrt{3}}\right)$ and $x \in \left(0 , \frac{1}{\sqrt{3}}\right)$
• $f \left(x\right)$ is concave downward on $x \in \left(- \frac{1}{\sqrt{3}} , 0\right)$ and $x \in \left(\frac{1}{\sqrt{3}} , 0\right)$.

• $f \left(x\right)$ has inflection points at $x = \pm \frac{1}{\sqrt{3}}$

FYI Here's a plot of $f \left(x\right)$ graph{1/x-3x^3 [-1.5, 1.5, -5, 5]}