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Calculus Examples
f(x)=-3x4-8x3-6x2+1f(x)=−3x4−8x3−6x2+1
Step 1
Step 1.1
By the Sum Rule, the derivative of -3x4-8x3-6x2+1−3x4−8x3−6x2+1 with respect to xx is ddx[-3x4]+ddx[-8x3]+ddx[-6x2]+ddx[1]ddx[−3x4]+ddx[−8x3]+ddx[−6x2]+ddx[1].
ddx[-3x4]+ddx[-8x3]+ddx[-6x2]+ddx[1]ddx[−3x4]+ddx[−8x3]+ddx[−6x2]+ddx[1]
Step 1.2
Evaluate ddx[-3x4]ddx[−3x4].
Step 1.2.1
Since -3−3 is constant with respect to xx, the derivative of -3x4−3x4 with respect to xx is -3ddx[x4]−3ddx[x4].
-3ddx[x4]+ddx[-8x3]+ddx[-6x2]+ddx[1]−3ddx[x4]+ddx[−8x3]+ddx[−6x2]+ddx[1]
Step 1.2.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=4n=4.
-3(4x3)+ddx[-8x3]+ddx[-6x2]+ddx[1]−3(4x3)+ddx[−8x3]+ddx[−6x2]+ddx[1]
Step 1.2.3
Multiply 44 by -3−3.
-12x3+ddx[-8x3]+ddx[-6x2]+ddx[1]−12x3+ddx[−8x3]+ddx[−6x2]+ddx[1]
-12x3+ddx[-8x3]+ddx[-6x2]+ddx[1]−12x3+ddx[−8x3]+ddx[−6x2]+ddx[1]
Step 1.3
Evaluate ddx[-8x3]ddx[−8x3].
Step 1.3.1
Since -8−8 is constant with respect to xx, the derivative of -8x3−8x3 with respect to xx is -8ddx[x3]−8ddx[x3].
-12x3-8ddx[x3]+ddx[-6x2]+ddx[1]−12x3−8ddx[x3]+ddx[−6x2]+ddx[1]
Step 1.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
-12x3-8(3x2)+ddx[-6x2]+ddx[1]
Step 1.3.3
Multiply 3 by -8.
-12x3-24x2+ddx[-6x2]+ddx[1]
-12x3-24x2+ddx[-6x2]+ddx[1]
Step 1.4
Evaluate ddx[-6x2].
Step 1.4.1
Since -6 is constant with respect to x, the derivative of -6x2 with respect to x is -6ddx[x2].
-12x3-24x2-6ddx[x2]+ddx[1]
Step 1.4.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
-12x3-24x2-6(2x)+ddx[1]
Step 1.4.3
Multiply 2 by -6.
-12x3-24x2-12x+ddx[1]
-12x3-24x2-12x+ddx[1]
Step 1.5
Differentiate using the Constant Rule.
Step 1.5.1
Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.
-12x3-24x2-12x+0
Step 1.5.2
Add -12x3-24x2-12x and 0.
-12x3-24x2-12x
-12x3-24x2-12x
-12x3-24x2-12x
Step 2
Step 2.1
By the Sum Rule, the derivative of -12x3-24x2-12x with respect to x is ddx[-12x3]+ddx[-24x2]+ddx[-12x].
f′′(x)=ddx(-12x3)+ddx(-24x2)+ddx(-12x)
Step 2.2
Evaluate ddx[-12x3].
Step 2.2.1
Since -12 is constant with respect to x, the derivative of -12x3 with respect to x is -12ddx[x3].
f′′(x)=-12ddxx3+ddx(-24x2)+ddx(-12x)
Step 2.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
f′′(x)=-12(3x2)+ddx(-24x2)+ddx(-12x)
Step 2.2.3
Multiply 3 by -12.
f′′(x)=-36x2+ddx(-24x2)+ddx(-12x)
f′′(x)=-36x2+ddx(-24x2)+ddx(-12x)
Step 2.3
Evaluate ddx[-24x2].
Step 2.3.1
Since -24 is constant with respect to x, the derivative of -24x2 with respect to x is -24ddx[x2].
f′′(x)=-36x2-24ddxx2+ddx(-12x)
Step 2.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
f′′(x)=-36x2-24(2x)+ddx(-12x)
Step 2.3.3
Multiply 2 by -24.
f′′(x)=-36x2-48x+ddx(-12x)
f′′(x)=-36x2-48x+ddx(-12x)
Step 2.4
Evaluate ddx[-12x].
Step 2.4.1
Since -12 is constant with respect to x, the derivative of -12x with respect to x is -12ddx[x].
f′′(x)=-36x2-48x-12ddxx
Step 2.4.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
f′′(x)=-36x2-48x-12⋅1
Step 2.4.3
Multiply -12 by 1.
f′′(x)=-36x2-48x-12
f′′(x)=-36x2-48x-12
f′′(x)=-36x2-48x-12
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
-12x3-24x2-12x=0
Step 4
Step 4.1
Find the first derivative.
Step 4.1.1
By the Sum Rule, the derivative of -3x4-8x3-6x2+1 with respect to x is ddx[-3x4]+ddx[-8x3]+ddx[-6x2]+ddx[1].
ddx[-3x4]+ddx[-8x3]+ddx[-6x2]+ddx[1]
Step 4.1.2
Evaluate ddx[-3x4].
Step 4.1.2.1
Since -3 is constant with respect to x, the derivative of -3x4 with respect to x is -3ddx[x4].
-3ddx[x4]+ddx[-8x3]+ddx[-6x2]+ddx[1]
Step 4.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=4.
-3(4x3)+ddx[-8x3]+ddx[-6x2]+ddx[1]
Step 4.1.2.3
Multiply 4 by -3.
-12x3+ddx[-8x3]+ddx[-6x2]+ddx[1]
-12x3+ddx[-8x3]+ddx[-6x2]+ddx[1]
Step 4.1.3
Evaluate ddx[-8x3].
Step 4.1.3.1
Since -8 is constant with respect to x, the derivative of -8x3 with respect to x is -8ddx[x3].
-12x3-8ddx[x3]+ddx[-6x2]+ddx[1]
Step 4.1.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
-12x3-8(3x2)+ddx[-6x2]+ddx[1]
Step 4.1.3.3
Multiply 3 by -8.
-12x3-24x2+ddx[-6x2]+ddx[1]
-12x3-24x2+ddx[-6x2]+ddx[1]
Step 4.1.4
Evaluate ddx[-6x2].
Step 4.1.4.1
Since -6 is constant with respect to x, the derivative of -6x2 with respect to x is -6ddx[x2].
-12x3-24x2-6ddx[x2]+ddx[1]
Step 4.1.4.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
-12x3-24x2-6(2x)+ddx[1]
Step 4.1.4.3
Multiply 2 by -6.
-12x3-24x2-12x+ddx[1]
-12x3-24x2-12x+ddx[1]
Step 4.1.5
Differentiate using the Constant Rule.
Step 4.1.5.1
Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.
-12x3-24x2-12x+0
Step 4.1.5.2
Add -12x3-24x2-12x and 0.
f′(x)=-12x3-24x2-12x
f′(x)=-12x3-24x2-12x
f′(x)=-12x3-24x2-12x
Step 4.2
The first derivative of f(x) with respect to x is -12x3-24x2-12x.
-12x3-24x2-12x
-12x3-24x2-12x
Step 5
Step 5.1
Set the first derivative equal to 0.
-12x3-24x2-12x=0
Step 5.2
Factor the left side of the equation.
Step 5.2.1
Factor -12x out of -12x3-24x2-12x.
Step 5.2.1.1
Factor -12x out of -12x3.
-12x⋅x2-24x2-12x=0
Step 5.2.1.2
Factor -12x out of -24x2.
-12x⋅x2-12x(2x)-12x=0
Step 5.2.1.3
Factor -12x out of -12x.
-12x⋅x2-12x(2x)-12x⋅1=0
Step 5.2.1.4
Factor -12x out of -12x(x2)-12x(2x).
-12x(x2+2x)-12x⋅1=0
Step 5.2.1.5
Factor -12x out of -12x(x2+2x)-12x(1).
-12x(x2+2x+1)=0
-12x(x2+2x+1)=0
Step 5.2.2
Factor using the perfect square rule.
Step 5.2.2.1
Rewrite 1 as 12.
-12x(x2+2x+12)=0
Step 5.2.2.2
Check that the middle term is two times the product of the numbers being squared in the first term and third term.
2x=2⋅x⋅1
Step 5.2.2.3
Rewrite the polynomial.
-12x(x2+2⋅x⋅1+12)=0
Step 5.2.2.4
Factor using the perfect square trinomial rule a2+2ab+b2=(a+b)2, where a=x and b=1.
-12x(x+1)2=0
-12x(x+1)2=0
-12x(x+1)2=0
Step 5.3
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
x=0
(x+1)2=0
Step 5.4
Set x equal to 0.
x=0
Step 5.5
Set (x+1)2 equal to 0 and solve for x.
Step 5.5.1
Set (x+1)2 equal to 0.
(x+1)2=0
Step 5.5.2
Solve (x+1)2=0 for x.
Step 5.5.2.1
Set the x+1 equal to 0.
x+1=0
Step 5.5.2.2
Subtract 1 from both sides of the equation.
x=-1
x=-1
x=-1
Step 5.6
The final solution is all the values that make -12x(x+1)2=0 true.
x=0,-1
x=0,-1
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=0,-1
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.
-36(0)2-48⋅0-12
Step 9
Step 9.1
Simplify each term.
Step 9.1.1
Raising 0 to any positive power yields 0.
-36⋅0-48⋅0-12
Step 9.1.2
Multiply -36 by 0.
0-48⋅0-12
Step 9.1.3
Multiply -48 by 0.
0+0-12
0+0-12
Step 9.2
Simplify by adding and subtracting.
Step 9.2.1
Add 0 and 0.
0-12
Step 9.2.2
Subtract 12 from 0.
-12
-12
-12
Step 10
x=0 is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.
x=0 is a local maximum
Step 11
Step 11.1
Replace the variable x with 0 in the expression.
f(0)=-3(0)4-8(0)3-6(0)2+1
Step 11.2
Simplify the result.
Step 11.2.1
Simplify each term.
Step 11.2.1.1
Raising 0 to any positive power yields 0.
f(0)=-3⋅0-8(0)3-6(0)2+1
Step 11.2.1.2
Multiply -3 by 0.
f(0)=0-8(0)3-6(0)2+1
Step 11.2.1.3
Raising 0 to any positive power yields 0.
f(0)=0-8⋅0-6(0)2+1
Step 11.2.1.4
Multiply -8 by 0.
f(0)=0+0-6(0)2+1
Step 11.2.1.5
Raising 0 to any positive power yields 0.
f(0)=0+0-6⋅0+1
Step 11.2.1.6
Multiply -6 by 0.
f(0)=0+0+0+1
f(0)=0+0+0+1
Step 11.2.2
Simplify by adding numbers.
Step 11.2.2.1
Add 0 and 0.
f(0)=0+0+1
Step 11.2.2.2
Add 0 and 0.
f(0)=0+1
Step 11.2.2.3
Add 0 and 1.
f(0)=1
f(0)=1
Step 11.2.3
The final answer is 1.
y=1
y=1
y=1
Step 12
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.
-36(-1)2-48⋅-1-12
Step 13
Step 13.1
Simplify each term.
Step 13.1.1
Raise -1 to the power of 2.
-36⋅1-48⋅-1-12
Step 13.1.2
Multiply -36 by 1.
-36-48⋅-1-12
Step 13.1.3
Multiply -48 by -1.
-36+48-12
-36+48-12
Step 13.2
Simplify by adding and subtracting.
Step 13.2.1
Add -36 and 48.
12-12
Step 13.2.2
Subtract 12 from 12.
0
0
0
Step 14
Step 14.1
Split (-∞,∞) into separate intervals around the x values that make the first derivative 0 or undefined.
(-∞,-1)∪(-1,0)∪(0,∞)
Step 14.2
Substitute any number, such as -4, from the interval (-∞,-1) in the first derivative -12x3-24x2-12x to check if the result is negative or positive.
Step 14.2.1
Replace the variable x with -4 in the expression.
f′(-4)=-12(-4)3-24(-4)2-12⋅-4
Step 14.2.2
Simplify the result.
Step 14.2.2.1
Simplify each term.
Step 14.2.2.1.1
Raise -4 to the power of 3.
f′(-4)=-12⋅-64-24(-4)2-12⋅-4
Step 14.2.2.1.2
Multiply -12 by -64.
f′(-4)=768-24(-4)2-12⋅-4
Step 14.2.2.1.3
Raise -4 to the power of 2.
f′(-4)=768-24⋅16-12⋅-4
Step 14.2.2.1.4
Multiply -24 by 16.
f′(-4)=768-384-12⋅-4
Step 14.2.2.1.5
Multiply -12 by -4.
f′(-4)=768-384+48
f′(-4)=768-384+48
Step 14.2.2.2
Simplify by adding and subtracting.
Step 14.2.2.2.1
Subtract 384 from 768.
f′(-4)=384+48
Step 14.2.2.2.2
Add 384 and 48.
f′(-4)=432
f′(-4)=432
Step 14.2.2.3
The final answer is 432.
432
432
432
Step 14.3
Substitute any number, such as -0.5, from the interval (-1,0) in the first derivative -12x3-24x2-12x to check if the result is negative or positive.
Step 14.3.1
Replace the variable x with -0.5 in the expression.
f′(-0.5)=-12(-0.5)3-24(-0.5)2-12⋅-0.5
Step 14.3.2
Simplify the result.
Step 14.3.2.1
Simplify each term.
Step 14.3.2.1.1
Raise -0.5 to the power of 3.
f′(-0.5)=-12⋅-0.125-24(-0.5)2-12⋅-0.5
Step 14.3.2.1.2
Multiply -12 by -0.125.
f′(-0.5)=1.5-24(-0.5)2-12⋅-0.5
Step 14.3.2.1.3
Raise -0.5 to the power of 2.
f′(-0.5)=1.5-24⋅0.25-12⋅-0.5
Step 14.3.2.1.4
Multiply -24 by 0.25.
f′(-0.5)=1.5-6-12⋅-0.5
Step 14.3.2.1.5
Multiply -12 by -0.5.
f′(-0.5)=1.5-6+6
f′(-0.5)=1.5-6+6
Step 14.3.2.2
Simplify by adding and subtracting.
Step 14.3.2.2.1
Subtract 6 from 1.5.
f′(-0.5)=-4.5+6
Step 14.3.2.2.2
Add -4.5 and 6.
f′(-0.5)=1.5
f′(-0.5)=1.5
Step 14.3.2.3
The final answer is 1.5.
1.5
1.5
1.5
Step 14.4
Substitute any number, such as 2, from the interval (0,∞) in the first derivative -12x3-24x2-12x to check if the result is negative or positive.
Step 14.4.1
Replace the variable x with 2 in the expression.
f′(2)=-12(2)3-24(2)2-12⋅2
Step 14.4.2
Simplify the result.
Step 14.4.2.1
Simplify each term.
Step 14.4.2.1.1
Raise 2 to the power of 3.
f′(2)=-12⋅8-24(2)2-12⋅2
Step 14.4.2.1.2
Multiply -12 by 8.
f′(2)=-96-24(2)2-12⋅2
Step 14.4.2.1.3
Raise 2 to the power of 2.
f′(2)=-96-24⋅4-12⋅2
Step 14.4.2.1.4
Multiply -24 by 4.
f′(2)=-96-96-12⋅2
Step 14.4.2.1.5
Multiply -12 by 2.
f′(2)=-96-96-24
f′(2)=-96-96-24
Step 14.4.2.2
Simplify by subtracting numbers.
Step 14.4.2.2.1
Subtract 96 from -96.
f′(2)=-192-24
Step 14.4.2.2.2
Subtract 24 from -192.
f′(2)=-216
f′(2)=-216
Step 14.4.2.3
The final answer is -216.
-216
-216
-216
Step 14.5
Since the first derivative did not change signs around x=-1, this is not a local maximum or minimum.
Not a local maximum or minimum
Step 14.6
Since the first derivative changed signs from positive to negative around x=0, then x=0 is a local maximum.
x=0 is a local maximum
x=0 is a local maximum
Step 15