Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=x^2+320/x ; (0,infinity)
f(x)=x2+320xf(x)=x2+320x ; (0,)
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+320x with respect to x is ddx[x2]+ddx[320x].
ddx[x2]+ddx[320x]
Step 1.1.1.1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
2x+ddx[320x]
2x+ddx[320x]
Step 1.1.1.2
Evaluate ddx[320x].
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Step 1.1.1.2.1
Since 320 is constant with respect to x, the derivative of 320x with respect to x is 320ddx[1x].
2x+320ddx[1x]
Step 1.1.1.2.2
Rewrite 1x as x-1.
2x+320ddx[x-1]
Step 1.1.1.2.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=-1.
2x+320(-x-2)
Step 1.1.1.2.4
Multiply -1 by 320.
2x-320x-2
2x-320x-2
Step 1.1.1.3
Simplify.
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Step 1.1.1.3.1
Rewrite the expression using the negative exponent rule b-n=1bn.
2x-3201x2
Step 1.1.1.3.2
Combine terms.
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Step 1.1.1.3.2.1
Combine -320 and 1x2.
2x+-320x2
Step 1.1.1.3.2.2
Move the negative in front of the fraction.
f(x)=2x-320x2
f(x)=2x-320x2
f(x)=2x-320x2
f(x)=2x-320x2
Step 1.1.2
The first derivative of f(x) with respect to x is 2x-320x2.
2x-320x2
2x-320x2
Step 1.2
Set the first derivative equal to 0 then solve the equation 2x-320x2=0.
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Step 1.2.1
Set the first derivative equal to 0.
2x-320x2=0
Step 1.2.2
Find the LCD of the terms in the equation.
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Step 1.2.2.1
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
1,x2,1
Step 1.2.2.2
The LCM of one and any expression is the expression.
x2
x2
Step 1.2.3
Multiply each term in 2x-320x2=0 by x2 to eliminate the fractions.
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Step 1.2.3.1
Multiply each term in 2x-320x2=0 by x2.
2xx2-320x2x2=0x2
Step 1.2.3.2
Simplify the left side.
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Step 1.2.3.2.1
Simplify each term.
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Step 1.2.3.2.1.1
Multiply x by x2 by adding the exponents.
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Step 1.2.3.2.1.1.1
Move x2.
2(x2x)-320x2x2=0x2
Step 1.2.3.2.1.1.2
Multiply x2 by x.
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Step 1.2.3.2.1.1.2.1
Raise x to the power of 1.
2(x2x1)-320x2x2=0x2
Step 1.2.3.2.1.1.2.2
Use the power rule aman=am+n to combine exponents.
2x2+1-320x2x2=0x2
2x2+1-320x2x2=0x2
Step 1.2.3.2.1.1.3
Add 2 and 1.
2x3-320x2x2=0x2
2x3-320x2x2=0x2
Step 1.2.3.2.1.2
Cancel the common factor of x2.
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Step 1.2.3.2.1.2.1
Move the leading negative in -320x2 into the numerator.
2x3+-320x2x2=0x2
Step 1.2.3.2.1.2.2
Cancel the common factor.
2x3+-320x2x2=0x2
Step 1.2.3.2.1.2.3
Rewrite the expression.
2x3-320=0x2
2x3-320=0x2
2x3-320=0x2
2x3-320=0x2
Step 1.2.3.3
Simplify the right side.
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Step 1.2.3.3.1
Multiply 0 by x2.
2x3-320=0
2x3-320=0
2x3-320=0
Step 1.2.4
Solve the equation.
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Step 1.2.4.1
Add 320 to both sides of the equation.
2x3=320
Step 1.2.4.2
Subtract 320 from both sides of the equation.
2x3-320=0
Step 1.2.4.3
Factor 2 out of 2x3-320.
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Step 1.2.4.3.1
Factor 2 out of 2x3.
2(x3)-320=0
Step 1.2.4.3.2
Factor 2 out of -320.
2x3+2-160=0
Step 1.2.4.3.3
Factor 2 out of 2x3+2-160.
2(x3-160)=0
2(x3-160)=0
Step 1.2.4.4
Divide each term in 2(x3-160)=0 by 2 and simplify.
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Step 1.2.4.4.1
Divide each term in 2(x3-160)=0 by 2.
2(x3-160)2=02
Step 1.2.4.4.2
Simplify the left side.
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Step 1.2.4.4.2.1
Cancel the common factor of 2.
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Step 1.2.4.4.2.1.1
Cancel the common factor.
2(x3-160)2=02
Step 1.2.4.4.2.1.2
Divide x3-160 by 1.
x3-160=02
x3-160=02
x3-160=02
Step 1.2.4.4.3
Simplify the right side.
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Step 1.2.4.4.3.1
Divide 0 by 2.
x3-160=0
x3-160=0
x3-160=0
Step 1.2.4.5
Add 160 to both sides of the equation.
x3=160
Step 1.2.4.6
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=3160
Step 1.2.4.7
Simplify 3160.
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Step 1.2.4.7.1
Rewrite 160 as 2320.
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Step 1.2.4.7.1.1
Factor 8 out of 160.
x=38(20)
Step 1.2.4.7.1.2
Rewrite 8 as 23.
x=32320
x=32320
Step 1.2.4.7.2
Pull terms out from under the radical.
x=2320
x=2320
x=2320
x=2320
Step 1.3
Find the values where the derivative is undefined.
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Step 1.3.1
Set the denominator in 320x2 equal to 0 to find where the expression is undefined.
x2=0
Step 1.3.2
Solve for x.
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Step 1.3.2.1
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=±0
Step 1.3.2.2
Simplify ±0.
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Step 1.3.2.2.1
Rewrite 0 as 02.
x=±02
Step 1.3.2.2.2
Pull terms out from under the radical, assuming positive real numbers.
x=±0
Step 1.3.2.2.3
Plus or minus 0 is 0.
x=0
x=0
x=0
x=0
Step 1.4
Evaluate x2+320x at each x value where the derivative is 0 or undefined.
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Step 1.4.1
Evaluate at x=2320.
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Step 1.4.1.1
Substitute 2320 for x.
(2320)2+3202320
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 2320.
223202+3202320
Step 1.4.1.2.1.2
Raise 2 to the power of 2.
43202+3202320
Step 1.4.1.2.1.3
Rewrite 3202 as 3202.
43202+3202320
Step 1.4.1.2.1.4
Raise 20 to the power of 2.
43400+3202320
Step 1.4.1.2.1.5
Rewrite 400 as 2350.
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Step 1.4.1.2.1.5.1
Factor 8 out of 400.
438(50)+3202320
Step 1.4.1.2.1.5.2
Rewrite 8 as 23.
432350+3202320
432350+3202320
Step 1.4.1.2.1.6
Pull terms out from under the radical.
4(2350)+3202320
Step 1.4.1.2.1.7
Multiply 2 by 4.
8350+3202320
Step 1.4.1.2.1.8
Cancel the common factor of 320 and 2.
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Step 1.4.1.2.1.8.1
Factor 2 out of 320.
8350+21602320
Step 1.4.1.2.1.8.2
Cancel the common factors.
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Step 1.4.1.2.1.8.2.1
Factor 2 out of 2320.
8350+21602(320)
Step 1.4.1.2.1.8.2.2
Cancel the common factor.
8350+21602320
Step 1.4.1.2.1.8.2.3
Rewrite the expression.
8350+160320
8350+160320
8350+160320
Step 1.4.1.2.1.9
Multiply 160320 by 32023202.
8350+16032032023202
Step 1.4.1.2.1.10
Combine and simplify the denominator.
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Step 1.4.1.2.1.10.1
Multiply 160320 by 32023202.
8350+16032023203202
Step 1.4.1.2.1.10.2
Raise 320 to the power of 1.
8350+160320232013202
Step 1.4.1.2.1.10.3
Use the power rule aman=am+n to combine exponents.
8350+16032023201+2
Step 1.4.1.2.1.10.4
Add 1 and 2.
8350+16032023203
Step 1.4.1.2.1.10.5
Rewrite 3203 as 20.
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Step 1.4.1.2.1.10.5.1
Use nax=axn to rewrite 320 as 2013.
8350+1603202(2013)3
Step 1.4.1.2.1.10.5.2
Apply the power rule and multiply exponents, (am)n=amn.
8350+160320220133
Step 1.4.1.2.1.10.5.3
Combine 13 and 3.
8350+16032022033
Step 1.4.1.2.1.10.5.4
Cancel the common factor of 3.
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Step 1.4.1.2.1.10.5.4.1
Cancel the common factor.
8350+16032022033
Step 1.4.1.2.1.10.5.4.2
Rewrite the expression.
8350+1603202201
8350+1603202201
Step 1.4.1.2.1.10.5.5
Evaluate the exponent.
8350+160320220
8350+160320220
8350+160320220
Step 1.4.1.2.1.11
Cancel the common factor of 160 and 20.
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Step 1.4.1.2.1.11.1
Factor 20 out of 1603202.
8350+20(83202)20
Step 1.4.1.2.1.11.2
Cancel the common factors.
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Step 1.4.1.2.1.11.2.1
Factor 20 out of 20.
8350+20(83202)20(1)
Step 1.4.1.2.1.11.2.2
Cancel the common factor.
8350+20(83202)201
Step 1.4.1.2.1.11.2.3
Rewrite the expression.
8350+832021
Step 1.4.1.2.1.11.2.4
Divide 83202 by 1.
8350+83202
8350+83202
8350+83202
Step 1.4.1.2.1.12
Rewrite 3202 as 3202.
8350+83202
Step 1.4.1.2.1.13
Raise 20 to the power of 2.
8350+83400
Step 1.4.1.2.1.14
Rewrite 400 as 2350.
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Step 1.4.1.2.1.14.1
Factor 8 out of 400.
8350+838(50)
Step 1.4.1.2.1.14.2
Rewrite 8 as 23.
8350+832350
8350+832350
Step 1.4.1.2.1.15
Pull terms out from under the radical.
8350+8(2350)
Step 1.4.1.2.1.16
Multiply 2 by 8.
8350+16350
8350+16350
Step 1.4.1.2.2
Add 8350 and 16350.
24350
24350
24350
Step 1.4.2
Evaluate at x=0.
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Step 1.4.2.1
Substitute 0 for x.
(0)2+3200
Step 1.4.2.2
The expression contains a division by 0. The expression is undefined.
Undefined
Undefined
Step 1.4.3
List all of the points.
(2320,24350)
(2320,24350)
(2320,24350)
Step 2
Use the first derivative test to determine which points can be maxima or minima.
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Step 2.1
Split (-,) into separate intervals around the x values that make the first derivative 0 or undefined.
(-,2320)(2320,)
Step 2.2
Substitute any number, such as -2, from the interval (-,2320) in the first derivative 2x-320x2 to check if the result is negative or positive.
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Step 2.2.1
Replace the variable x with -2 in the expression.
f(-2)=2(-2)-320(-2)2
Step 2.2.2
Simplify the result.
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Step 2.2.2.1
Simplify each term.
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Step 2.2.2.1.1
Multiply 2 by -2.
f(-2)=-4-320(-2)2
Step 2.2.2.1.2
Raise -2 to the power of 2.
f(-2)=-4-3204
Step 2.2.2.1.3
Divide 320 by 4.
f(-2)=-4-180
Step 2.2.2.1.4
Multiply -1 by 80.
f(-2)=-4-80
f(-2)=-4-80
Step 2.2.2.2
Subtract 80 from -4.
f(-2)=-84
Step 2.2.2.3
The final answer is -84.
-84
-84
-84
Step 2.3
Substitute any number, such as 8, from the interval (2320,) in the first derivative 2x-320x2 to check if the result is negative or positive.
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Step 2.3.1
Replace the variable x with 8 in the expression.
f(8)=2(8)-320(8)2
Step 2.3.2
Simplify the result.
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Step 2.3.2.1
Simplify each term.
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Step 2.3.2.1.1
Multiply 2 by 8.
f(8)=16-320(8)2
Step 2.3.2.1.2
Raise 8 to the power of 2.
f(8)=16-32064
Step 2.3.2.1.3
Divide 320 by 64.
f(8)=16-15
Step 2.3.2.1.4
Multiply -1 by 5.
f(8)=16-5
f(8)=16-5
Step 2.3.2.2
Subtract 5 from 16.
f(8)=11
Step 2.3.2.3
The final answer is 11.
11
11
11
Step 2.4
Since the first derivative changed signs from negative to positive around x=2320, then x=2320 is a local minimum.
x=2320 is a local minimum
x=2320 is a local minimum
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.
No absolute maximum
Absolute Minimum: (2320,24350)
Step 4
 [x2  12  π  xdx ]