Write as a function.

Differentiate.

By the Sum Rule, the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Evaluate .

Since is constant with respect to , the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Multiply by .

By the Sum Rule, the derivative of with respect to is .

Evaluate .

Since is constant with respect to , the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Multiply by .

Differentiate using the Constant Rule.

Since is constant with respect to , the derivative of with respect to is .

Add and .

To find the local maximum and minimum values of the function, set the derivative equal to and solve.

Add to both sides of the equation.

Divide each term in by .

Cancel the common factor of .

Cancel the common factor.

Divide by .

Divide by .

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

Any root of is .

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the to find the first solution.

Next, use the negative value of the to find the second solution.

The complete solution is the result of both the positive and negative portions of the solution.

Evaluate the second derivative at . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

One to any power is one.

Multiply by .

is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

is a local minimum

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

One to any power is one.

Multiply by .

Subtract from .

The final answer is .

Evaluate the second derivative at . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

Raise to the power of .

Multiply by .

is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

is a local maximum

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Add and .

The final answer is .

These are the local extrema for .

is a local minima

is a local maxima

Find the Local Maxima and Minima x^5-5x