Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=x^2-x-3 on 0 , 3
f(x)=x2-x-3 on 0 , 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
Differentiate.
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Step 1.1.1.1.1
By the Sum Rule, the derivative of x2-x-3 with respect to x is ddx[x2]+ddx[-x]+ddx[-3].
ddx[x2]+ddx[-x]+ddx[-3]
Step 1.1.1.1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
2x+ddx[-x]+ddx[-3]
2x+ddx[-x]+ddx[-3]
Step 1.1.1.2
Evaluate ddx[-x].
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Step 1.1.1.2.1
Since -1 is constant with respect to x, the derivative of -x with respect to x is -ddx[x].
2x-ddx[x]+ddx[-3]
Step 1.1.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
2x-11+ddx[-3]
Step 1.1.1.2.3
Multiply -1 by 1.
2x-1+ddx[-3]
2x-1+ddx[-3]
Step 1.1.1.3
Differentiate using the Constant Rule.
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Step 1.1.1.3.1
Since -3 is constant with respect to x, the derivative of -3 with respect to x is 0.
2x-1+0
Step 1.1.1.3.2
Add 2x-1 and 0.
f(x)=2x-1
f(x)=2x-1
f(x)=2x-1
Step 1.1.2
The first derivative of f(x) with respect to x is 2x-1.
2x-1
2x-1
Step 1.2
Set the first derivative equal to 0 then solve the equation 2x-1=0.
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Step 1.2.1
Set the first derivative equal to 0.
2x-1=0
Step 1.2.2
Add 1 to both sides of the equation.
2x=1
Step 1.2.3
Divide each term in 2x=1 by 2 and simplify.
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Step 1.2.3.1
Divide each term in 2x=1 by 2.
2x2=12
Step 1.2.3.2
Simplify the left side.
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Step 1.2.3.2.1
Cancel the common factor of 2.
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Step 1.2.3.2.1.1
Cancel the common factor.
2x2=12
Step 1.2.3.2.1.2
Divide x by 1.
x=12
x=12
x=12
x=12
x=12
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 x2-x-3 at each x value where the derivative is 0 or undefined.
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Step 1.4.1
Evaluate at x=12.
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Step 1.4.1.1
Substitute 12 for x.
(12)2-(12)-3
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 12.
1222-(12)-3
Step 1.4.1.2.1.2
One to any power is one.
122-(12)-3
Step 1.4.1.2.1.3
Raise 2 to the power of 2.
14-12-3
14-12-3
Step 1.4.1.2.2
Find the common denominator.
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Step 1.4.1.2.2.1
Multiply 12 by 22.
14-(1222)-3
Step 1.4.1.2.2.2
Multiply 12 by 22.
14-222-3
Step 1.4.1.2.2.3
Write -3 as a fraction with denominator 1.
14-222+-31
Step 1.4.1.2.2.4
Multiply -31 by 44.
14-222+-3144
Step 1.4.1.2.2.5
Multiply -31 by 44.
14-222+-344
Step 1.4.1.2.2.6
Multiply 2 by 2.
14-24+-344
14-24+-344
Step 1.4.1.2.3
Combine the numerators over the common denominator.
1-2-344
Step 1.4.1.2.4
Simplify the expression.
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Step 1.4.1.2.4.1
Multiply -3 by 4.
1-2-124
Step 1.4.1.2.4.2
Subtract 2 from 1.
-1-124
Step 1.4.1.2.4.3
Subtract 12 from -1.
-134
Step 1.4.1.2.4.4
Move the negative in front of the fraction.
-134
-134
-134
-134
Step 1.4.2
List all of the points.
(12,-134)
(12,-134)
(12,-134)
Step 2
Evaluate at the included endpoints.
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Step 2.1
Evaluate at x=0.
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Step 2.1.1
Substitute 0 for x.
(0)2-(0)-3
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
Raising 0 to any positive power yields 0.
0-(0)-3
Step 2.1.2.1.2
Multiply -1 by 0.
0+0-3
0+0-3
Step 2.1.2.2
Simplify by adding and subtracting.
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Step 2.1.2.2.1
Add 0 and 0.
0-3
Step 2.1.2.2.2
Subtract 3 from 0.
-3
-3
-3
-3
Step 2.2
Evaluate at x=3.
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Step 2.2.1
Substitute 3 for x.
(3)2-(3)-3
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.
9-(3)-3
Step 2.2.2.1.2
Multiply -1 by 3.
9-3-3
9-3-3
Step 2.2.2.2
Simplify by subtracting numbers.
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Step 2.2.2.2.1
Subtract 3 from 9.
6-3
Step 2.2.2.2.2
Subtract 3 from 6.
3
3
3
3
Step 2.3
List all of the points.
(0,-3),(3,3)
(0,-3),(3,3)
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: (3,3)
Absolute Minimum: (12,-134)
Step 4
 [x2  12  π  xdx ]