Find the Maximum/Minimum Value y=5x^10

Since ${\displaystyle 5}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 5x^{10}}$ with respect to ${\displaystyle x}$ is ${\displaystyle 5\frac{d}{dx}\left[x^{10}\right]}$.

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=10}$.

Multiply ${\displaystyle 10}$ by ${\displaystyle 5}$.

Since ${\displaystyle 50}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 50x^9}$ with respect to ${\displaystyle x}$ is ${\displaystyle 50\frac{d}{dx}\left[x^9\right]}$.

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=9}$.

Multiply ${\displaystyle 9}$ by ${\displaystyle 50}$.

Find the first derivative.

The first derivative of ${\displaystyle f\left(x\right)}$ with respect to ${\displaystyle x}$ is ${\displaystyle 50x^9}$.

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

Divide each term in ${\displaystyle 50x^9=0}$ by ${\displaystyle 50}$ and simplify.

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

Simplify ${\displaystyle \sqrt[9]{0}}$.

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 ${\displaystyle 450}$ 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 ${\displaystyle 50x^9}$ 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 ${\displaystyle 50x^9}$ 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.