Calculus Examples
,
Step 1
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 : .
Step 2
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.
Step 3
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 .
Differentiate using the Constant Rule.
Since is constant with respect to , the derivative of with respect to is .
Add and .
The first derivative of with respect to is .
Step 4
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.
Step 5
The function is differentiable on because the derivative is continuous on .
The function is differentiable.
Step 6
satisfies the two conditions for the mean value theorem. It is continuous on and differentiable on .
is continuous on and differentiable on .
Step 7
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 .
Step 8
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 .
Step 9
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 in by and simplify.
Divide each term in by .
Simplify the left side.
Cancel the common factor of .
Cancel the common factor.
Divide by .
Simplify the right side.
Cancel the common factor of and .
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.
Step 10
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
Step 11