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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
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
Since is constant with respect to , the derivative of with respect to is .
Step 3.1.2
The derivative of with respect to is .
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 .
is continuous on and differentiable on .
Step 7
Step 7.1
Replace the variable with in the expression.
Step 7.2
Simplify the result.
Step 7.2.1
The exact value of is .
Step 7.2.2
Multiply by .
Step 7.2.3
The final answer is .
Step 8
Step 8.1
Simplify.
Step 8.1.1
Multiply by .
Step 8.1.2
Multiply by .
Step 8.1.3
Reduce the expression by cancelling the common factors.
Step 8.1.3.1
Factor out of .
Step 8.1.3.2
Factor out of .
Step 8.1.3.3
Factor out of .
Step 8.1.3.4
Factor out of .
Step 8.1.3.5
Factor out of .
Step 8.1.3.6
Factor out of .
Step 8.1.3.7
Cancel the common factor.
Step 8.1.3.8
Rewrite the expression.
Step 8.1.4
Add and .
Step 8.1.5
Add and .
Step 8.1.6
Divide by .
Step 8.2
Divide each term in by and simplify.
Step 8.2.1
Divide each term in by .
Step 8.2.2
Simplify the left side.
Step 8.2.2.1
Cancel the common factor of .
Step 8.2.2.1.1
Cancel the common factor.
Step 8.2.2.1.2
Divide by .
Step 8.2.3
Simplify the right side.
Step 8.2.3.1
Divide by .
Step 8.3
Take the inverse cosine of both sides of the equation to extract from inside the cosine.
Step 8.4
Simplify the right side.
Step 8.4.1
The exact value of is .
Step 8.5
The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from to find the solution in the fourth quadrant.
Step 8.6
Simplify .
Step 8.6.1
To write as a fraction with a common denominator, multiply by .
Step 8.6.2
Combine fractions.
Step 8.6.2.1
Combine and .
Step 8.6.2.2
Combine the numerators over the common denominator.
Step 8.6.3
Simplify the numerator.
Step 8.6.3.1
Multiply by .
Step 8.6.3.2
Subtract from .
Step 8.7
Find the period of .
Step 8.7.1
The period of the function can be calculated using .
Step 8.7.2
Replace with in the formula for period.
Step 8.7.3
The absolute value is the distance between a number and zero. The distance between and is .
Step 8.7.4
Divide by .
Step 8.8
The period of the function is so values will repeat every radians in both directions.
, for any integer
Step 8.9
Consolidate the answers.
, for any integer
, for any integer
Step 9
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 10