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 .

Differentiate using the Constant Rule.

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

Add and .

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 .

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 .

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

Factor out of .

Factor out of .

Factor out of .

Factor out of .

Rewrite as .

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, where and .

Remove unnecessary parentheses.

Divide each term in by .

Cancel the common factor of .

Cancel the common factor.

Divide by .

Divide by .

Set the factor equal to .

Subtract from both sides of the equation.

Set the factor equal to .

Add to both sides of the equation.

The solution is the result of and .

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.

Simplify each term.

Raising to any positive power yields .

Multiply by .

Subtract from .

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.

Raising to any positive power yields .

Raising to any positive power yields .

Multiply by .

Simplify by adding zeros.

Add and .

Add and .

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.

Simplify each term.

Raise to the power of .

Multiply by .

Subtract from .

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.

Raise to the power of .

Raise to the power of .

Multiply by .

Simplify by adding and subtracting.

Subtract from .

Add and .

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.

Simplify each term.

Raise to the power of .

Multiply by .

Subtract from .

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.

Raise to the power of .

Raise to the power of .

Multiply by .

Simplify by adding and subtracting.

Subtract from .

Add and .

The final answer is .

These are the local extrema for .

is a local maxima

is a local minima

is a local minima

Find the Local Maxima and Minima f(x)=x^4-8x^2+1