Calculus Examples

Find the Local Maxima and Minima x^3-3x
x3-3x
Step 1
Write x3-3x as a function.
f(x)=x3-3x
Step 2
Find the first derivative of the function.
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Step 2.1
Differentiate.
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Step 2.1.1
By the Sum Rule, the derivative of x3-3x with respect to x is ddx[x3]+ddx[-3x].
ddx[x3]+ddx[-3x]
Step 2.1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
3x2+ddx[-3x]
3x2+ddx[-3x]
Step 2.2
Evaluate ddx[-3x].
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Step 2.2.1
Since -3 is constant with respect to x, the derivative of -3x with respect to x is -3ddx[x].
3x2-3ddx[x]
Step 2.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
3x2-31
Step 2.2.3
Multiply -3 by 1.
3x2-3
3x2-3
3x2-3
Step 3
Find the second derivative of the function.
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Step 3.1
By the Sum Rule, the derivative of 3x2-3 with respect to x is ddx[3x2]+ddx[-3].
f′′(x)=ddx(3x2)+ddx(-3)
Step 3.2
Evaluate ddx[3x2].
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Step 3.2.1
Since 3 is constant with respect to x, the derivative of 3x2 with respect to x is 3ddx[x2].
f′′(x)=3ddx(x2)+ddx(-3)
Step 3.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
f′′(x)=3(2x)+ddx(-3)
Step 3.2.3
Multiply 2 by 3.
f′′(x)=6x+ddx(-3)
f′′(x)=6x+ddx(-3)
Step 3.3
Differentiate using the Constant Rule.
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Step 3.3.1
Since -3 is constant with respect to x, the derivative of -3 with respect to x is 0.
f′′(x)=6x+0
Step 3.3.2
Add 6x and 0.
f′′(x)=6x
f′′(x)=6x
f′′(x)=6x
Step 4
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
3x2-3=0
Step 5
Find the first derivative.
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Step 5.1
Find the first derivative.
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Step 5.1.1
Differentiate.
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Step 5.1.1.1
By the Sum Rule, the derivative of x3-3x with respect to x is ddx[x3]+ddx[-3x].
ddx[x3]+ddx[-3x]
Step 5.1.1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
3x2+ddx[-3x]
3x2+ddx[-3x]
Step 5.1.2
Evaluate ddx[-3x].
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Step 5.1.2.1
Since -3 is constant with respect to x, the derivative of -3x with respect to x is -3ddx[x].
3x2-3ddx[x]
Step 5.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
3x2-31
Step 5.1.2.3
Multiply -3 by 1.
f(x)=3x2-3
f(x)=3x2-3
f(x)=3x2-3
Step 5.2
The first derivative of f(x) with respect to x is 3x2-3.
3x2-3
3x2-3
Step 6
Set the first derivative equal to 0 then solve the equation 3x2-3=0.
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Step 6.1
Set the first derivative equal to 0.
3x2-3=0
Step 6.2
Add 3 to both sides of the equation.
3x2=3
Step 6.3
Divide each term in 3x2=3 by 3 and simplify.
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Step 6.3.1
Divide each term in 3x2=3 by 3.
3x23=33
Step 6.3.2
Simplify the left side.
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Step 6.3.2.1
Cancel the common factor of 3.
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Step 6.3.2.1.1
Cancel the common factor.
3x23=33
Step 6.3.2.1.2
Divide x2 by 1.
x2=33
x2=33
x2=33
Step 6.3.3
Simplify the right side.
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Step 6.3.3.1
Divide 3 by 3.
x2=1
x2=1
x2=1
Step 6.4
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=±1
Step 6.5
Any root of 1 is 1.
x=±1
Step 6.6
The complete solution is the result of both the positive and negative portions of the solution.
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Step 6.6.1
First, use the positive value of the ± to find the first solution.
x=1
Step 6.6.2
Next, use the negative value of the ± to find the second solution.
x=-1
Step 6.6.3
The complete solution is the result of both the positive and negative portions of the solution.
x=1,-1
x=1,-1
x=1,-1
Step 7
Find the values where the derivative is undefined.
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Step 7.1
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 8
Critical points to evaluate.
x=1,-1
Step 9
Evaluate the second derivative at x=1. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
6(1)
Step 10
Multiply 6 by 1.
6
Step 11
x=1 is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
x=1 is a local minimum
Step 12
Find the y-value when x=1.
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Step 12.1
Replace the variable x with 1 in the expression.
f(1)=(1)3-31
Step 12.2
Simplify the result.
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Step 12.2.1
Simplify each term.
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Step 12.2.1.1
One to any power is one.
f(1)=1-31
Step 12.2.1.2
Multiply -3 by 1.
f(1)=1-3
f(1)=1-3
Step 12.2.2
Subtract 3 from 1.
f(1)=-2
Step 12.2.3
The final answer is -2.
y=-2
y=-2
y=-2
Step 13
Evaluate the second derivative at x=-1. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
6(-1)
Step 14
Multiply 6 by -1.
-6
Step 15
x=-1 is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.
x=-1 is a local maximum
Step 16
Find the y-value when x=-1.
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Step 16.1
Replace the variable x with -1 in the expression.
f(-1)=(-1)3-3-1
Step 16.2
Simplify the result.
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Step 16.2.1
Simplify each term.
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Step 16.2.1.1
Raise -1 to the power of 3.
f(-1)=-1-3-1
Step 16.2.1.2
Multiply -3 by -1.
f(-1)=-1+3
f(-1)=-1+3
Step 16.2.2
Add -1 and 3.
f(-1)=2
Step 16.2.3
The final answer is 2.
y=2
y=2
y=2
Step 17
These are the local extrema for f(x)=x3-3x.
(1,-2) is a local minima
(-1,2) is a local maxima
Step 18
image of graph
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