Calculus Examples

Find the Absolute Max and Min over the Interval
f(x)=5x2-3x-1f(x)=5x23x1 , [-1,3][1,3]
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
By the Sum Rule, the derivative of 5x2-3x-15x23x1 with respect to xx is ddx[5x2]+ddx[-3x]+ddx[-1]ddx[5x2]+ddx[3x]+ddx[1].
ddx[5x2]+ddx[-3x]+ddx[-1]ddx[5x2]+ddx[3x]+ddx[1]
Step 1.1.1.2
Evaluate ddx[5x2]ddx[5x2].
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Step 1.1.1.2.1
Since 55 is constant with respect to xx, the derivative of 5x25x2 with respect to xx is 5ddx[x2]5ddx[x2].
5ddx[x2]+ddx[-3x]+ddx[-1]5ddx[x2]+ddx[3x]+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.
5(2x)+ddx[-3x]+ddx[-1]5(2x)+ddx[3x]+ddx[1]
Step 1.1.1.2.3
Multiply 22 by 55.
10x+ddx[-3x]+ddx[-1]10x+ddx[3x]+ddx[1]
10x+ddx[-3x]+ddx[-1]10x+ddx[3x]+ddx[1]
Step 1.1.1.3
Evaluate ddx[-3x]ddx[3x].
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Step 1.1.1.3.1
Since -33 is constant with respect to xx, the derivative of -3x3x with respect to xx is -3ddx[x]3ddx[x].
10x-3ddx[x]+ddx[-1]10x3ddx[x]+ddx[1]
Step 1.1.1.3.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=1n=1.
10x-31+ddx[-1]10x31+ddx[1]
Step 1.1.1.3.3
Multiply -33 by 11.
10x-3+ddx[-1]10x3+ddx[1]
10x-3+ddx[-1]
Step 1.1.1.4
Differentiate using the Constant Rule.
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Step 1.1.1.4.1
Since -1 is constant with respect to x, the derivative of -1 with respect to x is 0.
10x-3+0
Step 1.1.1.4.2
Add 10x-3 and 0.
f(x)=10x-3
f(x)=10x-3
f(x)=10x-3
Step 1.1.2
The first derivative of f(x) with respect to x is 10x-3.
10x-3
10x-3
Step 1.2
Set the first derivative equal to 0 then solve the equation 10x-3=0.
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Step 1.2.1
Set the first derivative equal to 0.
10x-3=0
Step 1.2.2
Add 3 to both sides of the equation.
10x=3
Step 1.2.3
Divide each term in 10x=3 by 10 and simplify.
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Step 1.2.3.1
Divide each term in 10x=3 by 10.
10x10=310
Step 1.2.3.2
Simplify the left side.
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Step 1.2.3.2.1
Cancel the common factor of 10.
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Step 1.2.3.2.1.1
Cancel the common factor.
10x10=310
Step 1.2.3.2.1.2
Divide x by 1.
x=310
x=310
x=310
x=310
x=310
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 5x2-3x-1 at each x value where the derivative is 0 or undefined.
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Step 1.4.1
Evaluate at x=310.
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Step 1.4.1.1
Substitute 310 for x.
5(310)2-3(310)-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
Apply the product rule to 310.
532102-3(310)-1
Step 1.4.1.2.1.2
Raise 3 to the power of 2.
59102-3(310)-1
Step 1.4.1.2.1.3
Raise 10 to the power of 2.
5(9100)-3(310)-1
Step 1.4.1.2.1.4
Cancel the common factor of 5.
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Step 1.4.1.2.1.4.1
Factor 5 out of 100.
595(20)-3(310)-1
Step 1.4.1.2.1.4.2
Cancel the common factor.
59520-3(310)-1
Step 1.4.1.2.1.4.3
Rewrite the expression.
920-3(310)-1
920-3(310)-1
Step 1.4.1.2.1.5
Multiply -3(310).
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Step 1.4.1.2.1.5.1
Combine -3 and 310.
920+-3310-1
Step 1.4.1.2.1.5.2
Multiply -3 by 3.
920+-910-1
920+-910-1
Step 1.4.1.2.1.6
Move the negative in front of the fraction.
920-910-1
920-910-1
Step 1.4.1.2.2
Find the common denominator.
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Step 1.4.1.2.2.1
Multiply 910 by 22.
920-(91022)-1
Step 1.4.1.2.2.2
Multiply 910 by 22.
920-92102-1
Step 1.4.1.2.2.3
Write -1 as a fraction with denominator 1.
920-92102+-11
Step 1.4.1.2.2.4
Multiply -11 by 2020.
920-92102+-112020
Step 1.4.1.2.2.5
Multiply -11 by 2020.
920-92102+-12020
Step 1.4.1.2.2.6
Reorder the factors of 102.
920-92210+-12020
Step 1.4.1.2.2.7
Multiply 2 by 10.
920-9220+-12020
920-9220+-12020
Step 1.4.1.2.3
Combine the numerators over the common denominator.
9-92-12020
Step 1.4.1.2.4
Simplify each term.
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Step 1.4.1.2.4.1
Multiply -9 by 2.
9-18-12020
Step 1.4.1.2.4.2
Multiply -1 by 20.
9-18-2020
9-18-2020
Step 1.4.1.2.5
Simplify the expression.
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Step 1.4.1.2.5.1
Subtract 18 from 9.
-9-2020
Step 1.4.1.2.5.2
Subtract 20 from -9.
-2920
Step 1.4.1.2.5.3
Move the negative in front of the fraction.
-2920
-2920
-2920
-2920
Step 1.4.2
List all of the points.
(310,-2920)
(310,-2920)
(310,-2920)
Step 2
Evaluate at the included endpoints.
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Step 2.1
Evaluate at x=-1.
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Step 2.1.1
Substitute -1 for x.
5(-1)2-3-1-1
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
Raise -1 to the power of 2.
51-3-1-1
Step 2.1.2.1.2
Multiply 5 by 1.
5-3-1-1
Step 2.1.2.1.3
Multiply -3 by -1.
5+3-1
5+3-1
Step 2.1.2.2
Simplify by adding and subtracting.
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Step 2.1.2.2.1
Add 5 and 3.
8-1
Step 2.1.2.2.2
Subtract 1 from 8.
7
7
7
7
Step 2.2
Evaluate at x=3.
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Step 2.2.1
Substitute 3 for x.
5(3)2-33-1
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 3 to the power of 2.
59-33-1
Step 2.2.2.1.2
Multiply 5 by 9.
45-33-1
Step 2.2.2.1.3
Multiply -3 by 3.
45-9-1
45-9-1
Step 2.2.2.2
Simplify by subtracting numbers.
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Step 2.2.2.2.1
Subtract 9 from 45.
36-1
Step 2.2.2.2.2
Subtract 1 from 36.
35
35
35
35
Step 2.3
List all of the points.
(-1,7),(3,35)
(-1,7),(3,35)
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: (3,35)
Absolute Minimum: (310,-2920)
Step 4
Enter YOUR Problem
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 [x2  12  π  xdx ] 
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