Calculus Examples

Find the second derivative.
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Find the first derivative.
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By the Sum Rule, the derivative of with respect to is .
Differentiate using the Power Rule which states that is where .
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 to get .
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 to get .
Since is constant with respect to , the derivative of with respect to is .
Add and to get .
Find the second 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 to get .
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 to get .
Since is constant with respect to , the derivative of with respect to is .
Add and to get .
The derivative of with respect to is .
Set the second derivative equal to then solve the equation .
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Factor out of .
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Factor out of .
Factor out of .
Factor out of .
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 to get .
Divide by to get .
Set equal to and solve for .
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Set the factor equal to .
Since does not contain the variable to solve for, move it to the right side of the equation by subtracting from both sides.
The solution is the result of and .
Find the points where the second derivative is .
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Substitute in to find the value of .
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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 .
Raising to any positive power yields .
Remove parentheses around .
Raising to any positive power yields .
Multiply by to get .
Multiply by to get .
Simplify by adding zeros.
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Add and to get .
Add and to get .
Add and to get .
The final answer is .
The point found by substituting in is . This point can be an inflection point.
Substitute in to find the value of .
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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 to get .
Raise to the power of to get .
Multiply by to get .
Multiply by to get .
Simplify by adding and subtracting.
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Subtract from to get .
Add and to get .
Add and to get .
The final answer is .
The point found by substituting in is . This point can be an inflection point.
Determine the points that could be inflection points.
Split into intervals around the points that could potentially be inflection points.
Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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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 to get .
Multiply by to get .
Multiply by to get .
Subtract from to get .
The final answer is .
At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .
Increasing on since
Increasing on since
Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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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 to get .
Multiply by to get .
Multiply by to get .
Subtract from to get .
The final answer is .
At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .
Increasing on since
Increasing on since
Replace the variable with in the expression.
Simplify the result.
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Simplify each term.
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Apply the product rule to .
Raise to the power of to get .
Multiply by to get .
Apply the product rule to .
One to any power is one.
Raise to the power of to get .
Cancel the common factor of .
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Write as a fraction with denominator .
Factor out the greatest common factor .
Cancel the common factor.
Rewrite the expression.
Simplify.
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Multiply and to get .
Divide by to get .
Cancel the common factor of .
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Rewrite.
Write as a fraction with denominator .
Factor out the greatest common factor .
Cancel the common factor.
Rewrite the expression.
Reduce the expression by cancelling the common factors.
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Cancel the common factor.
Rewrite the expression.
Simplify.
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Multiply and to get .
Multiply by to get .
Divide by to get .
Subtract from to get .
The final answer is .
At , the second derivative is . Since this is negative, the second derivative is decreasing on the interval
Decreasing on since
Decreasing on since
Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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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 to get .
Multiply by to get .
Multiply by to get .
Subtract from to get .
The final answer is .
At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .
Increasing on since
Increasing on since
Replace the variable with in the expression.
Simplify the result.
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Simplify each term.
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Apply the product rule to .
Raise to the power of to get .
Multiply by to get .
Apply the product rule to .
One to any power is one.
Raise to the power of to get .
Cancel the common factor of .
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Write as a fraction with denominator .
Factor out the greatest common factor .
Cancel the common factor.
Rewrite the expression.
Simplify.
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Multiply and to get .
Divide by to get .
Cancel the common factor of .
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Rewrite.
Write as a fraction with denominator .
Factor out the greatest common factor .
Cancel the common factor.
Rewrite the expression.
Reduce the expression by cancelling the common factors.
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Cancel the common factor.
Rewrite the expression.
Simplify.
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Multiply and to get .
Multiply by to get .
Divide by to get .
Subtract from to get .
The final answer is .
At , the second derivative is . Since this is negative, the second derivative is decreasing on the interval
Decreasing on since
Decreasing on since
Replace the variable with in the expression.
Simplify the result.
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Simplify each term.
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Remove parentheses around .
Raise to the power of to get .
Multiply by to get .
Multiply by to get .
Add and to get .
The final answer is .
At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .
Increasing on since
Increasing on since
From to , the second derivative changes from increasing to decreasing. There is a valid inflection point at .
Inflection point at .
From to , the second derivative changes from decreasing to increasing. There is a valid inflection point at .
Inflection point at .
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 points in this case are .
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