# Calculus Examples

,

If is continuous on the interval and differentiable on , then at least one real number exists in the interval such that . The mean value theorem expresses the relationship between the slope of the tangent to the curve at and the slope of the line through the points and .

If is continuous on

and if differentiable on ,

then there exists at least one point, in : .

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.

is continuous on .

The function is continuous.

The function is continuous.

Find the first derivative.

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 .

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

Add and .

The first derivative of with respect to is .

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.

is continuous on .

The function is continuous.

The function is continuous.

The function is differentiable on because the derivative is continuous on .

The function is differentiable.

satisfies the two conditions for the mean value theorem. It is continuous on and differentiable on .

is continuous on and differentiable on .

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Multiply by .

Simplify by subtracting numbers.

Subtract from .

Subtract from .

The final answer is .

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

One to any power is one.

Multiply by .

Multiply by .

Simplify by adding and subtracting.

Add and .

Subtract from .

The final answer is .

Simplify .

Simplify the numerator.

Multiply by .

Add and .

Simplify the denominator.

Multiply by .

Add and .

Divide by .

Move all terms not containing to the right side of the equation.

Subtract from both sides of the equation.

Subtract from .

Divide each term by and simplify.

Divide each term in by .

Simplify the left side of the equation by cancelling the common factors.

Reduce the expression by cancelling the common factors.

Factor out of .

Move the negative one from the denominator of .

Rewrite as .

Simplify .

Reduce the expression by cancelling the common factors.

Factor out of .

Cancel the common factors.

Factor out of .

Cancel the common factor.

Rewrite the expression.

Move the negative in front of the fraction.

There is a tangent line found at parallel to the line that passes through the end points and .

There is a tangent line at parallel to the line that passes through the end points and