Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=x^3-3x^2+1 , [-1/2,4]
f(x)=x3-3x2+1f(x)=x33x2+1 , [-12,4][12,4]
Step 1
Find the critical points.
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Step 1.1
Find the first derivative.
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Step 1.1.1
Find the first derivative.
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Step 1.1.1.1
Differentiate.
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Step 1.1.1.1.1
By the Sum Rule, the derivative of x3-3x2+1x33x2+1 with respect to xx is ddx[x3]+ddx[-3x2]+ddx[1]ddx[x3]+ddx[3x2]+ddx[1].
ddx[x3]+ddx[-3x2]+ddx[1]ddx[x3]+ddx[3x2]+ddx[1]
Step 1.1.1.1.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=3n=3.
3x2+ddx[-3x2]+ddx[1]3x2+ddx[3x2]+ddx[1]
3x2+ddx[-3x2]+ddx[1]3x2+ddx[3x2]+ddx[1]
Step 1.1.1.2
Evaluate ddx[-3x2]ddx[3x2].
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Step 1.1.1.2.1
Since -33 is constant with respect to xx, the derivative of -3x23x2 with respect to xx is -3ddx[x2]3ddx[x2].
3x2-3ddx[x2]+ddx[1]3x23ddx[x2]+ddx[1]
Step 1.1.1.2.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=2n=2.
3x2-3(2x)+ddx[1]3x23(2x)+ddx[1]
Step 1.1.1.2.3
Multiply 22 by -33.
3x2-6x+ddx[1]3x26x+ddx[1]
3x2-6x+ddx[1]3x26x+ddx[1]
Step 1.1.1.3
Differentiate using the Constant Rule.
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Step 1.1.1.3.1
Since 11 is constant with respect to xx, the derivative of 11 with respect to xx is 00.
3x2-6x+03x26x+0
Step 1.1.1.3.2
Add 3x2-6x3x26x and 00.
f(x)=3x2-6x
f(x)=3x2-6x
f(x)=3x2-6x
Step 1.1.2
The first derivative of f(x) with respect to x is 3x2-6x.
3x2-6x
3x2-6x
Step 1.2
Set the first derivative equal to 0 then solve the equation 3x2-6x=0.
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Step 1.2.1
Set the first derivative equal to 0.
3x2-6x=0
Step 1.2.2
Factor 3x out of 3x2-6x.
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Step 1.2.2.1
Factor 3x out of 3x2.
3x(x)-6x=0
Step 1.2.2.2
Factor 3x out of -6x.
3x(x)+3x(-2)=0
Step 1.2.2.3
Factor 3x out of 3x(x)+3x(-2).
3x(x-2)=0
3x(x-2)=0
Step 1.2.3
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-2=0
Step 1.2.4
Set x equal to 0.
x=0
Step 1.2.5
Set x-2 equal to 0 and solve for x.
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Step 1.2.5.1
Set x-2 equal to 0.
x-2=0
Step 1.2.5.2
Add 2 to both sides of the equation.
x=2
x=2
Step 1.2.6
The final solution is all the values that make 3x(x-2)=0 true.
x=0,2
x=0,2
Step 1.3
Find the values where the derivative is undefined.
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Step 1.3.1
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 1.4
Evaluate x3-3x2+1 at each x value where the derivative is 0 or undefined.
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Step 1.4.1
Evaluate at x=0.
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Step 1.4.1.1
Substitute 0 for x.
(0)3-3(0)2+1
Step 1.4.1.2
Simplify.
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Step 1.4.1.2.1
Simplify each term.
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Step 1.4.1.2.1.1
Raising 0 to any positive power yields 0.
0-3(0)2+1
Step 1.4.1.2.1.2
Raising 0 to any positive power yields 0.
0-30+1
Step 1.4.1.2.1.3
Multiply -3 by 0.
0+0+1
0+0+1
Step 1.4.1.2.2
Simplify by adding numbers.
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Step 1.4.1.2.2.1
Add 0 and 0.
0+1
Step 1.4.1.2.2.2
Add 0 and 1.
1
1
1
1
Step 1.4.2
Evaluate at x=2.
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Step 1.4.2.1
Substitute 2 for x.
(2)3-3(2)2+1
Step 1.4.2.2
Simplify.
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Step 1.4.2.2.1
Simplify each term.
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Step 1.4.2.2.1.1
Raise 2 to the power of 3.
8-3(2)2+1
Step 1.4.2.2.1.2
Raise 2 to the power of 2.
8-34+1
Step 1.4.2.2.1.3
Multiply -3 by 4.
8-12+1
8-12+1
Step 1.4.2.2.2
Simplify by adding and subtracting.
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Step 1.4.2.2.2.1
Subtract 12 from 8.
-4+1
Step 1.4.2.2.2.2
Add -4 and 1.
-3
-3
-3
-3
Step 1.4.3
List all of the points.
(0,1),(2,-3)
(0,1),(2,-3)
(0,1),(2,-3)
Step 2
Evaluate at the included endpoints.
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Step 2.1
Evaluate at x=-12.
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Step 2.1.1
Substitute -12 for x.
(-12)3-3(-12)2+1
Step 2.1.2
Simplify.
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Step 2.1.2.1
Simplify each term.
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Step 2.1.2.1.1
Use the power rule (ab)n=anbn to distribute the exponent.
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Step 2.1.2.1.1.1
Apply the product rule to -12.
(-1)3(12)3-3(-12)2+1
Step 2.1.2.1.1.2
Apply the product rule to 12.
(-1)31323-3(-12)2+1
(-1)31323-3(-12)2+1
Step 2.1.2.1.2
Raise -1 to the power of 3.
-1323-3(-12)2+1
Step 2.1.2.1.3
One to any power is one.
-123-3(-12)2+1
Step 2.1.2.1.4
Raise 2 to the power of 3.
-18-3(-12)2+1
Step 2.1.2.1.5
Use the power rule (ab)n=anbn to distribute the exponent.
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Step 2.1.2.1.5.1
Apply the product rule to -12.
-18-3((-1)2(12)2)+1
Step 2.1.2.1.5.2
Apply the product rule to 12.
-18-3((-1)21222)+1
-18-3((-1)21222)+1
Step 2.1.2.1.6
Raise -1 to the power of 2.
-18-3(11222)+1
Step 2.1.2.1.7
Multiply 1222 by 1.
-18-31222+1
Step 2.1.2.1.8
One to any power is one.
-18-3122+1
Step 2.1.2.1.9
Raise 2 to the power of 2.
-18-3(14)+1
Step 2.1.2.1.10
Combine -3 and 14.
-18+-34+1
Step 2.1.2.1.11
Move the negative in front of the fraction.
-18-34+1
-18-34+1
Step 2.1.2.2
Find the common denominator.
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Step 2.1.2.2.1
Multiply 34 by 22.
-18-(3422)+1
Step 2.1.2.2.2
Multiply 34 by 22.
-18-3242+1
Step 2.1.2.2.3
Write 1 as a fraction with denominator 1.
-18-3242+11
Step 2.1.2.2.4
Multiply 11 by 88.
-18-3242+1188
Step 2.1.2.2.5
Multiply 11 by 88.
-18-3242+88
Step 2.1.2.2.6
Reorder the factors of 42.
-18-3224+88
Step 2.1.2.2.7
Multiply 2 by 4.
-18-328+88
-18-328+88
Step 2.1.2.3
Combine the numerators over the common denominator.
-1-32+88
Step 2.1.2.4
Simplify the expression.
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Step 2.1.2.4.1
Multiply -3 by 2.
-1-6+88
Step 2.1.2.4.2
Subtract 6 from -1.
-7+88
Step 2.1.2.4.3
Add -7 and 8.
18
18
18
18
Step 2.2
Evaluate at x=4.
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Step 2.2.1
Substitute 4 for x.
(4)3-3(4)2+1
Step 2.2.2
Simplify.
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Step 2.2.2.1
Simplify each term.
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Step 2.2.2.1.1
Raise 4 to the power of 3.
64-3(4)2+1
Step 2.2.2.1.2
Raise 4 to the power of 2.
64-316+1
Step 2.2.2.1.3
Multiply -3 by 16.
64-48+1
64-48+1
Step 2.2.2.2
Simplify by adding and subtracting.
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Step 2.2.2.2.1
Subtract 48 from 64.
16+1
Step 2.2.2.2.2
Add 16 and 1.
17
17
17
17
Step 2.3
List all of the points.
(-12,18),(4,17)
(-12,18),(4,17)
Step 3
Compare the f(x) values found for each value of x in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest f(x) value and the minimum will occur at the lowest f(x) value.
Absolute Maximum: (4,17)
Absolute Minimum: (2,-3)
Step 4
 [x2  12  π  xdx ]