Calculus Examples

Find the inflection points.
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Find the second derivative.
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Find the first derivative.
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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 .
Find the second derivative.
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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 .
The second derivative of with respect to is .
Set the second derivative equal to then solve the equation .
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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 .
Divide by .
Take the cube root of both sides of the equation to eliminate the exponent on the left side.
Simplify .
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Rewrite as .
Pull terms out from under the radical, assuming positive real numbers.
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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Raising to any positive power yields .
Multiply by .
The final answer is .
The point found by substituting in is . This point can be an inflection point.
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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Raise to the power of .
Multiply by .
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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Replace the variable with in the expression.
Simplify the result.
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Raise to the power of .
Multiply by .
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
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 .
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:
Create intervals around the inflection points and the undefined values.
Substitute any number from the interval into the second derivative and evaluate to determine the concavity.
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Replace the variable with in the expression.
Simplify the result.
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Raise to the power of .
Multiply by .
The final answer is .
The graph is concave up on the interval because is positive.
Concave up on since is positive
Concave up on since is positive
Substitute any number from the interval into the second derivative and evaluate to determine the concavity.
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Replace the variable with in the expression.
Simplify the result.
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Raise to the power of .
Multiply by .
The final answer is .
The graph is concave down on the interval because is negative.
Concave down on since is negative
Concave down on since is negative
The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.
Concave up on since is positive
Concave down on since is negative
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