Find the Maximum/Minimum Value f(x)=x^6-3.2

By the Sum Rule, the derivative of ${\displaystyle x^6-3.2}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[x^6\right]+\frac{d}{dx}[-3.2]}$.

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=6}$.

Since ${\displaystyle -3.2}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle -3.2}$ with respect to ${\displaystyle x}$ is ${\displaystyle 0}$.

Add ${\displaystyle 6x^5}$ and ${\displaystyle 0}$.

Since ${\displaystyle 6}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 6x^5}$ with respect to ${\displaystyle x}$ is ${\displaystyle 6\frac{d}{dx}\left[x^5\right]}$.

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=5}$.

Multiply ${\displaystyle 5}$ by ${\displaystyle 6}$.

Find the first derivative.

The first derivative of ${\displaystyle f\left(x\right)}$ with respect to ${\displaystyle x}$ is ${\displaystyle 6x^5}$.

Set the first derivative equal to ${\displaystyle 0}$.

Divide each term in ${\displaystyle 6x^5=0}$ by ${\displaystyle 6}$ and simplify.

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

Simplify ${\displaystyle \sqrt[5]{0}}$.

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Raising ${\displaystyle 0}$ to any positive power yields ${\displaystyle 0}$.

Multiply ${\displaystyle 30}$ by ${\displaystyle 0}$.

Split ${\displaystyle (-\infty ,\infty )}$ into separate intervals around the ${\displaystyle x}$ values that make the first derivative ${\displaystyle 0}$ or undefined.

Substitute any number, such as ${\displaystyle -2}$, from the interval ${\displaystyle (-\infty ,0)}$ in the first derivative ${\displaystyle 6x^5}$ to check if the result is negative or positive.

Substitute any number, such as ${\displaystyle 2}$, from the interval ${\displaystyle (0,\infty )}$ in the first derivative ${\displaystyle 6x^5}$ to check if the result is negative or positive.

Since the first derivative changed signs from negative to positive around ${\displaystyle x=0}$, then ${\displaystyle x=0}$ is a local minimum.