Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=x^3
f(x)=x3f(x)=x3
Step 1
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=3n=3.
3x23x2
Step 2
Find the second derivative of the function.
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Step 2.1
Since 33 is constant with respect to xx, the derivative of 3x23x2 with respect to xx is 3ddx[x2]3ddx[x2].
f′′(x)=3ddx(x2)
Step 2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
f′′(x)=3(2x)
Step 2.3
Multiply 2 by 3.
f′′(x)=6x
f′′(x)=6x
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
3x2=0
Step 4
Find the first derivative.
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Step 4.1
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
f(x)=3x2
Step 4.2
The first derivative of f(x) with respect to x is 3x2.
3x2
3x2
Step 5
Set the first derivative equal to 0 then solve the equation 3x2=0.
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Step 5.1
Set the first derivative equal to 0.
3x2=0
Step 5.2
Divide each term in 3x2=0 by 3 and simplify.
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Step 5.2.1
Divide each term in 3x2=0 by 3.
3x23=03
Step 5.2.2
Simplify the left side.
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Step 5.2.2.1
Cancel the common factor of 3.
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Step 5.2.2.1.1
Cancel the common factor.
3x23=03
Step 5.2.2.1.2
Divide x2 by 1.
x2=03
x2=03
x2=03
Step 5.2.3
Simplify the right side.
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Step 5.2.3.1
Divide 0 by 3.
x2=0
x2=0
x2=0
Step 5.3
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=±0
Step 5.4
Simplify ±0.
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Step 5.4.1
Rewrite 0 as 02.
x=±02
Step 5.4.2
Pull terms out from under the radical, assuming positive real numbers.
x=±0
Step 5.4.3
Plus or minus 0 is 0.
x=0
x=0
x=0
Step 6
Find the values where the derivative is undefined.
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Step 6.1
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 7
Critical points to evaluate.
x=0
Step 8
Evaluate the second derivative at x=0. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
6(0)
Step 9
Multiply 6 by 0.
0
Step 10
Since there is at least one point with 0 or undefined second derivative, apply the first derivative test.
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Step 10.1
Split (-,) into separate intervals around the x values that make the first derivative 0 or undefined.
(-,0)(0,)
Step 10.2
Substitute any number, such as -2, from the interval (-,0) in the first derivative 3x2 to check if the result is negative or positive.
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Step 10.2.1
Replace the variable x with -2 in the expression.
f(-2)=3(-2)2
Step 10.2.2
Simplify the result.
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Step 10.2.2.1
Raise -2 to the power of 2.
f(-2)=34
Step 10.2.2.2
Multiply 3 by 4.
f(-2)=12
Step 10.2.2.3
The final answer is 12.
12
12
12
Step 10.3
Substitute any number, such as 2, from the interval (0,) in the first derivative 3x2 to check if the result is negative or positive.
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Step 10.3.1
Replace the variable x with 2 in the expression.
f(2)=3(2)2
Step 10.3.2
Simplify the result.
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Step 10.3.2.1
Raise 2 to the power of 2.
f(2)=34
Step 10.3.2.2
Multiply 3 by 4.
f(2)=12
Step 10.3.2.3
The final answer is 12.
12
12
12
Step 10.4
Since the first derivative did not change signs around x=0, this is not a local maximum or minimum.
Not a local maximum or minimum
Step 10.5
No local maxima or minima found for f(x)=x3.
No local maxima or minima
No local maxima or minima
Step 11
 [x2  12  π  xdx ]