Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=x^3-3/2x^2
f(x)=x3-32x2f(x)=x332x2
Step 1
Find the first derivative of the function.
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Step 1.1
Differentiate.
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Step 1.1.1
By the Sum Rule, the derivative of x3-32x2x332x2 with respect to xx is ddx[x3]+ddx[-32x2]ddx[x3]+ddx[32x2].
ddx[x3]+ddx[-32x2]ddx[x3]+ddx[32x2]
Step 1.1.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=3n=3.
3x2+ddx[-32x2]3x2+ddx[32x2]
3x2+ddx[-32x2]3x2+ddx[32x2]
Step 1.2
Evaluate ddx[-32x2]ddx[32x2].
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Step 1.2.1
Since -3232 is constant with respect to xx, the derivative of -32x232x2 with respect to xx is -32ddx[x2]32ddx[x2].
3x2-32ddx[x2]3x232ddx[x2]
Step 1.2.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=2n=2.
3x2-32(2x)3x232(2x)
Step 1.2.3
Multiply 22 by -11.
3x2-2(32)x3x22(32)x
Step 1.2.4
Combine -22 and 3232.
3x2+-232x3x2+232x
Step 1.2.5
Multiply -22 by 33.
3x2+-62x3x2+62x
Step 1.2.6
Combine -6262 and xx.
3x2+-6x23x2+6x2
Step 1.2.7
Cancel the common factor of -66 and 22.
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Step 1.2.7.1
Factor 22 out of -6x6x.
3x2+2(-3x)23x2+2(3x)2
Step 1.2.7.2
Cancel the common factors.
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Step 1.2.7.2.1
Factor 22 out of 22.
3x2+2(-3x)2(1)3x2+2(3x)2(1)
Step 1.2.7.2.2
Cancel the common factor.
3x2+2(-3x)21
Step 1.2.7.2.3
Rewrite the expression.
3x2+-3x1
Step 1.2.7.2.4
Divide -3x by 1.
3x2-3x
3x2-3x
3x2-3x
3x2-3x
3x2-3x
Step 2
Find the second derivative of the function.
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Step 2.1
By the Sum Rule, the derivative of 3x2-3x with respect to x is ddx[3x2]+ddx[-3x].
f′′(x)=ddx(3x2)+ddx(-3x)
Step 2.2
Evaluate ddx[3x2].
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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(-3x)
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(-3x)
Step 2.2.3
Multiply 2 by 3.
f′′(x)=6x+ddx(-3x)
f′′(x)=6x+ddx(-3x)
Step 2.3
Evaluate ddx[-3x].
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Step 2.3.1
Since -3 is constant with respect to x, the derivative of -3x with respect to x is -3ddx[x].
f′′(x)=6x-3ddxx
Step 2.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
f′′(x)=6x-31
Step 2.3.3
Multiply -3 by 1.
f′′(x)=6x-3
f′′(x)=6x-3
f′′(x)=6x-3
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
3x2-3x=0
Step 4
Find the first derivative.
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Step 4.1
Find the first derivative.
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Step 4.1.1
Differentiate.
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Step 4.1.1.1
By the Sum Rule, the derivative of x3-32x2 with respect to x is ddx[x3]+ddx[-32x2].
ddx[x3]+ddx[-32x2]
Step 4.1.1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
3x2+ddx[-32x2]
3x2+ddx[-32x2]
Step 4.1.2
Evaluate ddx[-32x2].
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Step 4.1.2.1
Since -32 is constant with respect to x, the derivative of -32x2 with respect to x is -32ddx[x2].
3x2-32ddx[x2]
Step 4.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
3x2-32(2x)
Step 4.1.2.3
Multiply 2 by -1.
3x2-2(32)x
Step 4.1.2.4
Combine -2 and 32.
3x2+-232x
Step 4.1.2.5
Multiply -2 by 3.
3x2+-62x
Step 4.1.2.6
Combine -62 and x.
3x2+-6x2
Step 4.1.2.7
Cancel the common factor of -6 and 2.
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Step 4.1.2.7.1
Factor 2 out of -6x.
3x2+2(-3x)2
Step 4.1.2.7.2
Cancel the common factors.
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Step 4.1.2.7.2.1
Factor 2 out of 2.
3x2+2(-3x)2(1)
Step 4.1.2.7.2.2
Cancel the common factor.
3x2+2(-3x)21
Step 4.1.2.7.2.3
Rewrite the expression.
3x2+-3x1
Step 4.1.2.7.2.4
Divide -3x by 1.
f(x)=3x2-3x
f(x)=3x2-3x
f(x)=3x2-3x
f(x)=3x2-3x
f(x)=3x2-3x
Step 4.2
The first derivative of f(x) with respect to x is 3x2-3x.
3x2-3x
3x2-3x
Step 5
Set the first derivative equal to 0 then solve the equation 3x2-3x=0.
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Step 5.1
Set the first derivative equal to 0.
3x2-3x=0
Step 5.2
Factor 3x out of 3x2-3x.
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Step 5.2.1
Factor 3x out of 3x2.
3x(x)-3x=0
Step 5.2.2
Factor 3x out of -3x.
3x(x)+3x(-1)=0
Step 5.2.3
Factor 3x out of 3x(x)+3x(-1).
3x(x-1)=0
3x(x-1)=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=0
Step 5.4
Set x equal to 0.
x=0
Step 5.5
Set x-1 equal to 0 and solve for x.
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Step 5.5.1
Set x-1 equal to 0.
x-1=0
Step 5.5.2
Add 1 to both sides of the equation.
x=1
x=1
Step 5.6
The final solution is all the values that make 3x(x-1)=0 true.
x=0,1
x=0,1
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=0,1
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.
6(0)-3
Step 9
Evaluate the second derivative.
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Step 9.1
Multiply 6 by 0.
0-3
Step 9.2
Subtract 3 from 0.
-3
-3
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
Find the y-value when x=0.
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Step 11.1
Replace the variable x with 0 in the expression.
f(0)=(0)3-32(0)2
Step 11.2
Simplify the result.
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Step 11.2.1
Simplify each term.
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Step 11.2.1.1
Raising 0 to any positive power yields 0.
f(0)=0-32(0)2
Step 11.2.1.2
Raising 0 to any positive power yields 0.
f(0)=0-320
Step 11.2.1.3
Multiply -320.
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Step 11.2.1.3.1
Multiply 0 by -1.
f(0)=0+0(32)
Step 11.2.1.3.2
Multiply 0 by 32.
f(0)=0+0
f(0)=0+0
f(0)=0+0
Step 11.2.2
Add 0 and 0.
f(0)=0
Step 11.2.3
The final answer is 0.
y=0
y=0
y=0
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.
6(1)-3
Step 13
Evaluate the second derivative.
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Step 13.1
Multiply 6 by 1.
6-3
Step 13.2
Subtract 3 from 6.
3
3
Step 14
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 15
Find the y-value when x=1.
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Step 15.1
Replace the variable x with 1 in the expression.
f(1)=(1)3-32(1)2
Step 15.2
Simplify the result.
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Step 15.2.1
Simplify each term.
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Step 15.2.1.1
One to any power is one.
f(1)=1-32(1)2
Step 15.2.1.2
One to any power is one.
f(1)=1-321
Step 15.2.1.3
Multiply -1 by 1.
f(1)=1-32
f(1)=1-32
Step 15.2.2
Simplify the expression.
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Step 15.2.2.1
Write 1 as a fraction with a common denominator.
f(1)=22-32
Step 15.2.2.2
Combine the numerators over the common denominator.
f(1)=2-32
Step 15.2.2.3
Subtract 3 from 2.
f(1)=-12
Step 15.2.2.4
Move the negative in front of the fraction.
f(1)=-12
f(1)=-12
Step 15.2.3
The final answer is -12.
y=-12
y=-12
y=-12
Step 16
These are the local extrema for f(x)=x3-32x2.
(0,0) is a local maxima
(1,-12) is a local minima
Step 17
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