Calculus Examples

Find Where the Mean Value Theorem is Satisfied
,
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 : .
Check if is continuous.
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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 derivative.
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Find the first derivative.
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By the Sum Rule, the derivative of with respect to is .
Evaluate .
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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 .
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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 derivative of with respect to is .
Find if the derivative is continuous on .
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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 .
Evaluate from the interval .
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Replace the variable with in the expression.
Simplify the result.
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Simplify each term.
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Raise to the power of .
Multiply by .
Multiply by .
Simplify by subtracting numbers.
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Subtract from .
Subtract from .
The final answer is .
Evaluate from the interval .
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Replace the variable with in the expression.
Simplify the result.
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Simplify each term.
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Remove parentheses around .
One to any power is one.
Multiply by .
Multiply by .
Simplify by adding and subtracting.
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Add and .
Subtract from .
The final answer is .
Solve for . .
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Simplify the right side.
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Simplify the numerator.
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Multiply by .
Add and .
Simplify the denominator.
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Remove parentheses.
Multiply by .
Add and .
Divide by .
Move all terms not containing to the right side of the equation.
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Subtract from both sides of the equation.
Add and .
Divide each term by and simplify.
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Divide each term in by .
Reduce the expression by cancelling the common factors.
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Cancel the common factor.
Divide by .
Reduce the expression by cancelling the common factors.
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Factor out of .
Cancel the common factor.
Rewrite the expression.
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
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