# 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 to get .

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 get .

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

Add and to get .

The 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 to get .

Multiply by to get .

Multiply by to get .

Simplify by subtracting numbers.

Subtract from to get .

Subtract from to get .

The final answer is .

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses around .

One to any power is one.

Multiply by to get .

Multiply by to get .

Simplify by adding and subtracting.

Add and to get .

Subtract from to get .

The final answer is .

Simplify the right side.

Simplify the numerator.

Multiply by to get .

Subtract from to get .

Simplify the denominator.

Remove parentheses.

Multiply by to get .

Add and to get .

Divide by to get .

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

Since does not contain the variable to solve for, move it to the right side of the equation by subtracting from both sides.

Subtract from to get .

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 to get .

Simplify the right side of the equation.

Divide by to get .

Multiply by to get .

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