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Calculus Examples
f(x)=x3-12xf(x)=x3−12x
Step 1
Step 1.1
Differentiate.
Step 1.1.1
By the Sum Rule, the derivative of x3-12xx3−12x with respect to xx is ddx[x3]+ddx[-12x]ddx[x3]+ddx[−12x].
ddx[x3]+ddx[-12x]ddx[x3]+ddx[−12x]
Step 1.1.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=3n=3.
3x2+ddx[-12x]3x2+ddx[−12x]
3x2+ddx[-12x]3x2+ddx[−12x]
Step 1.2
Evaluate ddx[-12x]ddx[−12x].
Step 1.2.1
Since -12−12 is constant with respect to xx, the derivative of -12x−12x with respect to xx is -12ddx[x]−12ddx[x].
3x2-12ddx[x]3x2−12ddx[x]
Step 1.2.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
3x2-12⋅13x2−12⋅1
Step 1.2.3
Multiply -12−12 by 11.
3x2-123x2−12
3x2-123x2−12
3x2-123x2−12
Step 2
Step 2.1
By the Sum Rule, the derivative of 3x2-123x2−12 with respect to xx is ddx[3x2]+ddx[-12]ddx[3x2]+ddx[−12].
f′′(x)=ddx(3x2)+ddx(-12)
Step 2.2
Evaluate ddx[3x2].
Step 2.2.1
Since 3 is constant with respect to x, the derivative of 3x2 with respect to x is 3ddx[x2].
f′′(x)=3ddx(x2)+ddx(-12)
Step 2.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
f′′(x)=3(2x)+ddx(-12)
Step 2.2.3
Multiply 2 by 3.
f′′(x)=6x+ddx(-12)
f′′(x)=6x+ddx(-12)
Step 2.3
Differentiate using the Constant Rule.
Step 2.3.1
Since -12 is constant with respect to x, the derivative of -12 with respect to x is 0.
f′′(x)=6x+0
Step 2.3.2
Add 6x and 0.
f′′(x)=6x
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.
3x2-12=0
Step 4
Step 4.1
Find the first derivative.
Step 4.1.1
Differentiate.
Step 4.1.1.1
By the Sum Rule, the derivative of x3-12x with respect to x is ddx[x3]+ddx[-12x].
ddx[x3]+ddx[-12x]
Step 4.1.1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
3x2+ddx[-12x]
3x2+ddx[-12x]
Step 4.1.2
Evaluate ddx[-12x].
Step 4.1.2.1
Since -12 is constant with respect to x, the derivative of -12x with respect to x is -12ddx[x].
3x2-12ddx[x]
Step 4.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
3x2-12⋅1
Step 4.1.2.3
Multiply -12 by 1.
f′(x)=3x2-12
f′(x)=3x2-12
f′(x)=3x2-12
Step 4.2
The first derivative of f(x) with respect to x is 3x2-12.
3x2-12
3x2-12
Step 5
Step 5.1
Set the first derivative equal to 0.
3x2-12=0
Step 5.2
Add 12 to both sides of the equation.
3x2=12
Step 5.3
Divide each term in 3x2=12 by 3 and simplify.
Step 5.3.1
Divide each term in 3x2=12 by 3.
3x23=123
Step 5.3.2
Simplify the left side.
Step 5.3.2.1
Cancel the common factor of 3.
Step 5.3.2.1.1
Cancel the common factor.
3x23=123
Step 5.3.2.1.2
Divide x2 by 1.
x2=123
x2=123
x2=123
Step 5.3.3
Simplify the right side.
Step 5.3.3.1
Divide 12 by 3.
x2=4
x2=4
x2=4
Step 5.4
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=±√4
Step 5.5
Simplify ±√4.
Step 5.5.1
Rewrite 4 as 22.
x=±√22
Step 5.5.2
Pull terms out from under the radical, assuming positive real numbers.
x=±2
x=±2
Step 5.6
The complete solution is the result of both the positive and negative portions of the solution.
Step 5.6.1
First, use the positive value of the ± to find the first solution.
x=2
Step 5.6.2
Next, use the negative value of the ± to find the second solution.
x=-2
Step 5.6.3
The complete solution is the result of both the positive and negative portions of the solution.
x=2,-2
x=2,-2
x=2,-2
Step 6
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=2,-2
Step 8
Evaluate the second derivative at x=2. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
6(2)
Step 9
Multiply 6 by 2.
12
Step 10
x=2 is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
x=2 is a local minimum
Step 11
Step 11.1
Replace the variable x with 2 in the expression.
f(2)=(2)3-12⋅2
Step 11.2
Simplify the result.
Step 11.2.1
Simplify each term.
Step 11.2.1.1
Raise 2 to the power of 3.
f(2)=8-12⋅2
Step 11.2.1.2
Multiply -12 by 2.
f(2)=8-24
f(2)=8-24
Step 11.2.2
Subtract 24 from 8.
f(2)=-16
Step 11.2.3
The final answer is -16.
y=-16
y=-16
y=-16
Step 12
Evaluate the second derivative at x=-2. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
6(-2)
Step 13
Multiply 6 by -2.
-12
Step 14
x=-2 is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.
x=-2 is a local maximum
Step 15
Step 15.1
Replace the variable x with -2 in the expression.
f(-2)=(-2)3-12⋅-2
Step 15.2
Simplify the result.
Step 15.2.1
Simplify each term.
Step 15.2.1.1
Raise -2 to the power of 3.
f(-2)=-8-12⋅-2
Step 15.2.1.2
Multiply -12 by -2.
f(-2)=-8+24
f(-2)=-8+24
Step 15.2.2
Add -8 and 24.
f(-2)=16
Step 15.2.3
The final answer is 16.
y=16
y=16
y=16
Step 16
These are the local extrema for f(x)=x3-12x.
(2,-16) is a local minima
(-2,16) is a local maxima
Step 17