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Calculus Examples
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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
Step 2.1
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:
Step 2.2
is continuous on .
The function is continuous.
The function is continuous.
Step 3
Step 3.1
Find the first derivative.
Step 3.1.1
Differentiate.
Step 3.1.1.1
By the Sum Rule, the derivative of with respect to is .
Step 3.1.1.2
Differentiate using the Power Rule which states that is where .
Step 3.1.2
Evaluate .
Step 3.1.2.1
Since is constant with respect to , the derivative of with respect to is .
Step 3.1.2.2
Differentiate using the Power Rule which states that is where .
Step 3.1.2.3
Multiply by .
Step 3.2
The first derivative of with respect to is .
Step 4
Step 4.1
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:
Step 4.2
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 .
No solution
Step 7
Step 7.1
Replace the variable with in the expression.
Step 7.2
Simplify the result.
Step 7.2.1
Simplify each term.
Step 7.2.1.1
One to any power is one.
Step 7.2.1.2
Multiply by .
Step 7.2.2
Subtract from .
Step 7.2.3
The final answer is .
Step 8
Step 8.1
Replace the variable with in the expression.
Step 8.2
Simplify the result.
Step 8.2.1
Simplify each term.
Step 8.2.1.1
Raise to the power of .
Step 8.2.1.2
Multiply by .
Step 8.2.2
Add and .
Step 8.2.3
The final answer is .
Step 9
Step 9.1
Simplify .
Step 9.1.1
Simplify the numerator.
Step 9.1.1.1
Multiply by .
Step 9.1.1.2
Add and .
Step 9.1.2
Simplify the denominator.
Step 9.1.2.1
Multiply by .
Step 9.1.2.2
Subtract from .
Step 9.1.3
Divide by .
Step 9.2
Move all terms not containing to the right side of the equation.
Step 9.2.1
Add to both sides of the equation.
Step 9.2.2
Add and .
Step 9.3
Divide each term in by and simplify.
Step 9.3.1
Divide each term in by .
Step 9.3.2
Simplify the left side.
Step 9.3.2.1
Cancel the common factor of .
Step 9.3.2.1.1
Cancel the common factor.
Step 9.3.2.1.2
Divide by .
Step 9.4
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
Step 9.5
Simplify .
Step 9.5.1
Rewrite as .
Step 9.5.2
Multiply by .
Step 9.5.3
Combine and simplify the denominator.
Step 9.5.3.1
Multiply by .
Step 9.5.3.2
Raise to the power of .
Step 9.5.3.3
Raise to the power of .
Step 9.5.3.4
Use the power rule to combine exponents.
Step 9.5.3.5
Add and .
Step 9.5.3.6
Rewrite as .
Step 9.5.3.6.1
Use to rewrite as .
Step 9.5.3.6.2
Apply the power rule and multiply exponents, .
Step 9.5.3.6.3
Combine and .
Step 9.5.3.6.4
Cancel the common factor of .
Step 9.5.3.6.4.1
Cancel the common factor.
Step 9.5.3.6.4.2
Rewrite the expression.
Step 9.5.3.6.5
Evaluate the exponent.
Step 9.5.4
Simplify the numerator.
Step 9.5.4.1
Combine using the product rule for radicals.
Step 9.5.4.2
Multiply by .
Step 9.6
The complete solution is the result of both the positive and negative portions of the solution.
Step 9.6.1
First, use the positive value of the to find the first solution.
Step 9.6.2
Next, use the negative value of the to find the second solution.
Step 9.6.3
The complete solution is the result of both the positive and negative portions of the solution.
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
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 12