# 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.

Interval Notation:

Set-Builder Notation:

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.

Interval Notation:

Set-Builder Notation:

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.

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 .

Reduce the expression by cancelling the common factors.

Cancel the common factor.

Divide by .

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