Calculus Examples

Find the Inflection Points y=x-sin(x)
Step 1
Write as a function.
Step 2
Find the second derivative.
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Step 2.1
Find the first derivative.
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Step 2.1.1
Differentiate.
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Step 2.1.1.1
By the Sum Rule, the derivative of with respect to is .
Step 2.1.1.2
Differentiate using the Power Rule which states that is where .
Step 2.1.2
Evaluate .
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Step 2.1.2.1
Since is constant with respect to , the derivative of with respect to is .
Step 2.1.2.2
The derivative of with respect to is .
Step 2.2
Find the second derivative.
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Step 2.2.1
Differentiate.
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Step 2.2.1.1
By the Sum Rule, the derivative of with respect to is .
Step 2.2.1.2
Since is constant with respect to , the derivative of with respect to is .
Step 2.2.2
Evaluate .
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Step 2.2.2.1
Since is constant with respect to , the derivative of with respect to is .
Step 2.2.2.2
The derivative of with respect to is .
Step 2.2.2.3
Multiply by .
Step 2.2.2.4
Multiply by .
Step 2.2.3
Add and .
Step 2.3
The second derivative of with respect to is .
Step 3
Set the second derivative equal to then solve the equation .
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Step 3.1
Set the second derivative equal to .
Step 3.2
Take the inverse sine of both sides of the equation to extract from inside the sine.
Step 3.3
Simplify the right side.
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Step 3.3.1
The exact value of is .
Step 3.4
The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from to find the solution in the second quadrant.
Step 3.5
Subtract from .
Step 3.6
Find the period of .
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Step 3.6.1
The period of the function can be calculated using .
Step 3.6.2
Replace with in the formula for period.
Step 3.6.3
The absolute value is the distance between a number and zero. The distance between and is .
Step 3.6.4
Divide by .
Step 3.7
The period of the function is so values will repeat every radians in both directions.
, for any integer
Step 3.8
Consolidate the answers.
, for any integer
, for any integer
Step 4
The point found by substituting in is . This point can be an inflection point.
Step 5
Split into intervals around the points that could potentially be inflection points.
Step 6
Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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Step 6.1
Replace the variable with in the expression.
Step 6.2
The final answer is .
Step 6.3
At , the second derivative is . Since this is negative, the second derivative is decreasing on the interval
Decreasing on since
Decreasing on since
Step 7
Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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Step 7.1
Replace the variable with in the expression.
Step 7.2
The final answer is .
Step 7.3
At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .
Increasing on since
Increasing on since
Step 8
An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is .
Step 9