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Calculus Examples
f(x)=x2+320xf(x)=x2+320x ; (0,∞)
Step 1
Step 1.1
Find the first derivative.
Step 1.1.1
Find the first derivative.
Step 1.1.1.1
Differentiate.
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].
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.
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.
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.
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.
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.
Step 1.2.3.1
Multiply each term in 2x-320x2=0 by x2.
2x⋅x2-320x2x2=0x2
Step 1.2.3.2
Simplify the left side.
Step 1.2.3.2.1
Simplify each term.
Step 1.2.3.2.1.1
Multiply x by x2 by adding the exponents.
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.
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.
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.
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.
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.
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.
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.
Step 1.2.4.4.2.1
Cancel the common factor of 2.
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.
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=3√160
Step 1.2.4.7
Simplify 3√160.
Step 1.2.4.7.1
Rewrite 160 as 23⋅20.
Step 1.2.4.7.1.1
Factor 8 out of 160.
x=3√8(20)
Step 1.2.4.7.1.2
Rewrite 8 as 23.
x=3√23⋅20
x=3√23⋅20
Step 1.2.4.7.2
Pull terms out from under the radical.
x=23√20
x=23√20
x=23√20
x=23√20
Step 1.3
Find the values where the derivative is undefined.
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.
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.
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.
Step 1.4.1
Evaluate at x=23√20.
Step 1.4.1.1
Substitute 23√20 for x.
(23√20)2+32023√20
Step 1.4.1.2
Simplify.
Step 1.4.1.2.1
Simplify each term.
Step 1.4.1.2.1.1
Apply the product rule to 23√20.
223√202+32023√20
Step 1.4.1.2.1.2
Raise 2 to the power of 2.
43√202+32023√20
Step 1.4.1.2.1.3
Rewrite 3√202 as 3√202.
43√202+32023√20
Step 1.4.1.2.1.4
Raise 20 to the power of 2.
43√400+32023√20
Step 1.4.1.2.1.5
Rewrite 400 as 23⋅50.
Step 1.4.1.2.1.5.1
Factor 8 out of 400.
43√8(50)+32023√20
Step 1.4.1.2.1.5.2
Rewrite 8 as 23.
43√23⋅50+32023√20
43√23⋅50+32023√20
Step 1.4.1.2.1.6
Pull terms out from under the radical.
4(23√50)+32023√20
Step 1.4.1.2.1.7
Multiply 2 by 4.
83√50+32023√20
Step 1.4.1.2.1.8
Cancel the common factor of 320 and 2.
Step 1.4.1.2.1.8.1
Factor 2 out of 320.
83√50+2⋅16023√20
Step 1.4.1.2.1.8.2
Cancel the common factors.
Step 1.4.1.2.1.8.2.1
Factor 2 out of 23√20.
83√50+2⋅1602(3√20)
Step 1.4.1.2.1.8.2.2
Cancel the common factor.
83√50+2⋅16023√20
Step 1.4.1.2.1.8.2.3
Rewrite the expression.
83√50+1603√20
83√50+1603√20
83√50+1603√20
Step 1.4.1.2.1.9
Multiply 1603√20 by 3√2023√202.
83√50+1603√20⋅3√2023√202
Step 1.4.1.2.1.10
Combine and simplify the denominator.
Step 1.4.1.2.1.10.1
Multiply 1603√20 by 3√2023√202.
83√50+1603√2023√203√202
Step 1.4.1.2.1.10.2
Raise 3√20 to the power of 1.
83√50+1603√2023√2013√202
Step 1.4.1.2.1.10.3
Use the power rule aman=am+n to combine exponents.
83√50+1603√2023√201+2
Step 1.4.1.2.1.10.4
Add 1 and 2.
83√50+1603√2023√203
Step 1.4.1.2.1.10.5
Rewrite 3√203 as 20.
Step 1.4.1.2.1.10.5.1
Use n√ax=axn to rewrite 3√20 as 2013.
83√50+1603√202(2013)3
Step 1.4.1.2.1.10.5.2
Apply the power rule and multiply exponents, (am)n=amn.
83√50+1603√2022013⋅3
Step 1.4.1.2.1.10.5.3
Combine 13 and 3.
83√50+1603√2022033
Step 1.4.1.2.1.10.5.4
Cancel the common factor of 3.
Step 1.4.1.2.1.10.5.4.1
Cancel the common factor.
83√50+1603√2022033
Step 1.4.1.2.1.10.5.4.2
Rewrite the expression.
83√50+1603√202201
83√50+1603√202201
Step 1.4.1.2.1.10.5.5
Evaluate the exponent.
83√50+1603√20220
83√50+1603√20220
83√50+1603√20220
Step 1.4.1.2.1.11
Cancel the common factor of 160 and 20.
Step 1.4.1.2.1.11.1
Factor 20 out of 1603√202.
83√50+20(83√202)20
Step 1.4.1.2.1.11.2
Cancel the common factors.
Step 1.4.1.2.1.11.2.1
Factor 20 out of 20.
83√50+20(83√202)20(1)
Step 1.4.1.2.1.11.2.2
Cancel the common factor.
83√50+20(83√202)20⋅1
Step 1.4.1.2.1.11.2.3
Rewrite the expression.
83√50+83√2021
Step 1.4.1.2.1.11.2.4
Divide 83√202 by 1.
83√50+83√202
83√50+83√202
83√50+83√202
Step 1.4.1.2.1.12
Rewrite 3√202 as 3√202.
83√50+83√202
Step 1.4.1.2.1.13
Raise 20 to the power of 2.
83√50+83√400
Step 1.4.1.2.1.14
Rewrite 400 as 23⋅50.
Step 1.4.1.2.1.14.1
Factor 8 out of 400.
83√50+83√8(50)
Step 1.4.1.2.1.14.2
Rewrite 8 as 23.
83√50+83√23⋅50
83√50+83√23⋅50
Step 1.4.1.2.1.15
Pull terms out from under the radical.
83√50+8(23√50)
Step 1.4.1.2.1.16
Multiply 2 by 8.
83√50+163√50
83√50+163√50
Step 1.4.1.2.2
Add 83√50 and 163√50.
243√50
243√50
243√50
Step 1.4.2
Evaluate at x=0.
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.
(23√20,243√50)
(23√20,243√50)
(23√20,243√50)
Step 2
Step 2.1
Split (-∞,∞) into separate intervals around the x values that make the first derivative 0 or undefined.
(-∞,23√20)∪(23√20,∞)
Step 2.2
Substitute any number, such as -2, from the interval (-∞,23√20) in the first derivative 2x-320x2 to check if the result is negative or positive.
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.
Step 2.2.2.1
Simplify each term.
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)=-4-1⋅80
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 (23√20,∞) in the first derivative 2x-320x2 to check if the result is negative or positive.
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.
Step 2.3.2.1
Simplify each term.
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)=16-1⋅5
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=23√20, then x=23√20 is a local minimum.
x=23√20 is a local minimum
x=23√20 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: (23√20,243√50)
Step 4