Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=x^2-2x , (0,2)
,
Step 1
Find the critical points.
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Step 1.1
Find the first derivative.
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Step 1.1.1
Find the first derivative.
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Step 1.1.1.1
Differentiate.
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Step 1.1.1.1.1
By the Sum Rule, the derivative of with respect to is .
Step 1.1.1.1.2
Differentiate using the Power Rule which states that is where .
Step 1.1.1.2
Evaluate .
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Step 1.1.1.2.1
Since is constant with respect to , the derivative of with respect to is .
Step 1.1.1.2.2
Differentiate using the Power Rule which states that is where .
Step 1.1.1.2.3
Multiply by .
Step 1.1.2
The first derivative of with respect to is .
Step 1.2
Set the first derivative equal to then solve the equation .
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Step 1.2.1
Set the first derivative equal to .
Step 1.2.2
Add to both sides of the equation.
Step 1.2.3
Divide each term in by and simplify.
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Step 1.2.3.1
Divide each term in by .
Step 1.2.3.2
Simplify the left side.
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Step 1.2.3.2.1
Cancel the common factor of .
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Step 1.2.3.2.1.1
Cancel the common factor.
Step 1.2.3.2.1.2
Divide by .
Step 1.2.3.3
Simplify the right side.
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Step 1.2.3.3.1
Divide by .
Step 1.3
Find the values where the derivative is undefined.
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Step 1.3.1
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.
Step 1.4
Evaluate at each value where the derivative is or undefined.
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Step 1.4.1
Evaluate at .
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Step 1.4.1.1
Substitute for .
Step 1.4.1.2
Simplify.
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Step 1.4.1.2.1
Simplify each term.
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Step 1.4.1.2.1.1
One to any power is one.
Step 1.4.1.2.1.2
Multiply by .
Step 1.4.1.2.2
Subtract from .
Step 1.4.2
List all of the points.
Step 2
Use the first derivative test to determine which points can be maxima or minima.
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Step 2.1
Split into separate intervals around the values that make the first derivative or undefined.
Step 2.2
Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.
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Step 2.2.1
Replace the variable with in the expression.
Step 2.2.2
Simplify the result.
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Step 2.2.2.1
Multiply by .
Step 2.2.2.2
Subtract from .
Step 2.2.2.3
The final answer is .
Step 2.3
Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.
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Step 2.3.1
Replace the variable with in the expression.
Step 2.3.2
Simplify the result.
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Step 2.3.2.1
Multiply by .
Step 2.3.2.2
Subtract from .
Step 2.3.2.3
The final answer is .
Step 2.4
Since the first derivative changed signs from negative to positive around , then is a local minimum.
is a local minimum
is a local minimum
Step 3
Compare the values found for each value of in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest value and the minimum will occur at the lowest value.
No absolute maximum
Absolute Minimum:
Step 4