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Calculus Examples
f(x)=23√x-1+3f(x)=23√x−1+3 ; [-7,9][−7,9]
Step 1
Step 1.1
Find the first derivative.
Step 1.1.1
Find the first derivative.
Step 1.1.1.1
By the Sum Rule, the derivative of 23√x-1+323√x−1+3 with respect to xx is ddx[23√x-1]+ddx[3]ddx[23√x−1]+ddx[3].
f′(x)=ddx(23√x-1)+ddx(3)f'(x)=ddx(23√x−1)+ddx(3)
Step 1.1.1.2
Evaluate ddx[23√x-1]ddx[23√x−1].
Step 1.1.1.2.1
Use n√ax=axnn√ax=axn to rewrite 3√x-13√x−1 as (x-1)13(x−1)13.
f′(x)=ddx(2(x-1)13)+ddx(3)f'(x)=ddx(2(x−1)13)+ddx(3)
Step 1.1.1.2.2
Since 22 is constant with respect to xx, the derivative of 2(x-1)132(x−1)13 with respect to xx is 2ddx[(x-1)13]2ddx[(x−1)13].
f′(x)=2ddx((x-1)13)+ddx(3)f'(x)=2ddx((x−1)13)+ddx(3)
Step 1.1.1.2.3
Differentiate using the chain rule, which states that ddx[f(g(x))]ddx[f(g(x))] is f′(g(x))g′(x)f'(g(x))g'(x) where f(x)=x13f(x)=x13 and g(x)=x-1g(x)=x−1.
Step 1.1.1.2.3.1
To apply the Chain Rule, set uu as x-1x−1.
f′(x)=2(ddu(u13)ddx(x-1))+ddx(3)f'(x)=2(ddu(u13)ddx(x−1))+ddx(3)
Step 1.1.1.2.3.2
Differentiate using the Power Rule which states that ddu[un]ddu[un] is nun-1nun−1 where n=13n=13.
f′(x)=2(13u13-1ddx(x-1))+ddx(3)f'(x)=2(13u13−1ddx(x−1))+ddx(3)
Step 1.1.1.2.3.3
Replace all occurrences of uu with x-1x−1.
f′(x)=2(13⋅(x-1)13-1ddx(x-1))+ddx(3)f'(x)=2(13⋅(x−1)13−1ddx(x−1))+ddx(3)
f′(x)=2(13⋅(x-1)13-1ddx(x-1))+ddx(3)f'(x)=2(13⋅(x−1)13−1ddx(x−1))+ddx(3)
Step 1.1.1.2.4
By the Sum Rule, the derivative of x-1x−1 with respect to xx is ddx[x]+ddx[-1]ddx[x]+ddx[−1].
f′(x)=2(13⋅((x-1)13-1(ddx(x)+ddx(-1))))+ddx(3)f'(x)=2(13⋅((x−1)13−1(ddx(x)+ddx(−1))))+ddx(3)
Step 1.1.1.2.5
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
f′(x)=2(13⋅((x-1)13-1(1+ddx(-1))))+ddx(3)f'(x)=2(13⋅((x−1)13−1(1+ddx(−1))))+ddx(3)
Step 1.1.1.2.6
Since -1−1 is constant with respect to xx, the derivative of -1−1 with respect to xx is 00.
f′(x)=2(13⋅((x-1)13-1(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)13−1(1+0)))+ddx(3)
Step 1.1.1.2.7
To write -1−1 as a fraction with a common denominator, multiply by 3333.
f′(x)=2(13⋅((x-1)13-1⋅33(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)13−1⋅33(1+0)))+ddx(3)
Step 1.1.1.2.8
Combine -1−1 and 3333.
f′(x)=2(13⋅((x-1)13+-1⋅33(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)13+−1⋅33(1+0)))+ddx(3)
Step 1.1.1.2.9
Combine the numerators over the common denominator.
f′(x)=2(13⋅((x-1)1-1⋅33(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)1−1⋅33(1+0)))+ddx(3)
Step 1.1.1.2.10
Simplify the numerator.
Step 1.1.1.2.10.1
Multiply -1−1 by 33.
f′(x)=2(13⋅((x-1)1-33(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)1−33(1+0)))+ddx(3)
Step 1.1.1.2.10.2
Subtract 33 from 11.
f′(x)=2(13⋅((x-1)-23(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)−23(1+0)))+ddx(3)
f′(x)=2(13⋅((x-1)-23(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)−23(1+0)))+ddx(3)
Step 1.1.1.2.11
Move the negative in front of the fraction.
f′(x)=2(13⋅((x-1)-23(1+0)))+ddx(3)f'(x)=2(13⋅((x−1)−23(1+0)))+ddx(3)
Step 1.1.1.2.12
Add 11 and 00.
f′(x)=2(13⋅(x-1)-23⋅1)+ddx(3)f'(x)=2(13⋅(x−1)−23⋅1)+ddx(3)
Step 1.1.1.2.13
Combine 1313 and (x-1)-23(x−1)−23.
f′(x)=2((x-1)-233⋅1)+ddx(3)f'(x)=2((x−1)−233⋅1)+ddx(3)
Step 1.1.1.2.14
Multiply (x-1)-233(x−1)−233 by 11.
f′(x)=2((x-1)-233)+ddx(3)f'(x)=2((x−1)−233)+ddx(3)
Step 1.1.1.2.15
Move (x-1)-23(x−1)−23 to the denominator using the negative exponent rule b-n=1bnb−n=1bn.
f′(x)=2(13(x-1)23)+ddx(3)f'(x)=2(13(x−1)23)+ddx(3)
Step 1.1.1.2.16
Combine 22 and 13(x-1)2313(x−1)23.
f′(x)=23(x-1)23+ddx(3)f'(x)=23(x−1)23+ddx(3)
f′(x)=23(x-1)23+ddx(3)f'(x)=23(x−1)23+ddx(3)
Step 1.1.1.3
Differentiate using the Constant Rule.
Step 1.1.1.3.1
Since 33 is constant with respect to xx, the derivative of 33 with respect to xx is 00.
f′(x)=23(x-1)23+0f'(x)=23(x−1)23+0
Step 1.1.1.3.2
Add 23(x-1)2323(x−1)23 and 00.
f′(x)=23(x-1)23f'(x)=23(x−1)23
f′(x)=23(x-1)23f'(x)=23(x−1)23
f′(x)=23(x-1)23f'(x)=23(x−1)23
Step 1.1.2
The first derivative of f(x)f(x) with respect to xx is 23(x-1)2323(x−1)23.
23(x-1)2323(x−1)23
23(x-1)2323(x−1)23
Step 1.2
Set the first derivative equal to 00 then solve the equation 23(x-1)23=023(x−1)23=0.
Step 1.2.1
Set the first derivative equal to 00.
23(x-1)23=023(x−1)23=0
Step 1.2.2
Set the numerator equal to zero.
2=02=0
Step 1.2.3
Since 2≠02≠0, there are no solutions.
No solution
No solution
Step 1.3
Find the values where the derivative is undefined.
Step 1.3.1
Apply the rule xmn=n√xmxmn=n√xm to rewrite the exponentiation as a radical.
233√(x-1)2233√(x−1)2
Step 1.3.2
Set the denominator in 233√(x-1)2233√(x−1)2 equal to 00 to find where the expression is undefined.
33√(x-1)2=033√(x−1)2=0
Step 1.3.3
Solve for xx.
Step 1.3.3.1
To remove the radical on the left side of the equation, cube both sides of the equation.
(33√(x-1)2)3=03(33√(x−1)2)3=03
Step 1.3.3.2
Simplify each side of the equation.
Step 1.3.3.2.1
Use n√ax=axnn√ax=axn to rewrite 3√(x-1)23√(x−1)2 as (x-1)23(x−1)23.
(3(x-1)23)3=03(3(x−1)23)3=03
Step 1.3.3.2.2
Simplify the left side.
Step 1.3.3.2.2.1
Simplify (3(x-1)23)3(3(x−1)23)3.
Step 1.3.3.2.2.1.1
Apply the product rule to 3(x-1)233(x−1)23.
33((x-1)23)3=0333((x−1)23)3=03
Step 1.3.3.2.2.1.2
Raise 33 to the power of 33.
27((x-1)23)3=0327((x−1)23)3=03
Step 1.3.3.2.2.1.3
Multiply the exponents in ((x-1)23)3((x−1)23)3.
Step 1.3.3.2.2.1.3.1
Apply the power rule and multiply exponents, (am)n=amn(am)n=amn.
27(x-1)23⋅3=0327(x−1)23⋅3=03
Step 1.3.3.2.2.1.3.2
Cancel the common factor of 33.
Step 1.3.3.2.2.1.3.2.1
Cancel the common factor.
27(x-1)23⋅3=03
Step 1.3.3.2.2.1.3.2.2
Rewrite the expression.
27(x-1)2=03
27(x-1)2=03
27(x-1)2=03
27(x-1)2=03
27(x-1)2=03
Step 1.3.3.2.3
Simplify the right side.
Step 1.3.3.2.3.1
Raising 0 to any positive power yields 0.
27(x-1)2=0
27(x-1)2=0
27(x-1)2=0
Step 1.3.3.3
Solve for x.
Step 1.3.3.3.1
Divide each term in 27(x-1)2=0 by 27 and simplify.
Step 1.3.3.3.1.1
Divide each term in 27(x-1)2=0 by 27.
27(x-1)227=027
Step 1.3.3.3.1.2
Simplify the left side.
Step 1.3.3.3.1.2.1
Cancel the common factor of 27.
Step 1.3.3.3.1.2.1.1
Cancel the common factor.
27(x-1)227=027
Step 1.3.3.3.1.2.1.2
Divide (x-1)2 by 1.
(x-1)2=027
(x-1)2=027
(x-1)2=027
Step 1.3.3.3.1.3
Simplify the right side.
Step 1.3.3.3.1.3.1
Divide 0 by 27.
(x-1)2=0
(x-1)2=0
(x-1)2=0
Step 1.3.3.3.2
Set the x-1 equal to 0.
x-1=0
Step 1.3.3.3.3
Add 1 to both sides of the equation.
x=1
x=1
x=1
x=1
Step 1.4
Evaluate 23√x-1+3 at each x value where the derivative is 0 or undefined.
Step 1.4.1
Evaluate at x=1.
Step 1.4.1.1
Substitute 1 for x.
23√(1)-1+3
Step 1.4.1.2
Simplify.
Step 1.4.1.2.1
Simplify each term.
Step 1.4.1.2.1.1
Subtract 1 from 1.
23√0+3
Step 1.4.1.2.1.2
Rewrite 0 as 03.
23√03+3
Step 1.4.1.2.1.3
Pull terms out from under the radical, assuming real numbers.
2⋅0+3
Step 1.4.1.2.1.4
Multiply 2 by 0.
0+3
0+3
Step 1.4.1.2.2
Add 0 and 3.
3
3
3
Step 1.4.2
List all of the points.
(1,3)
(1,3)
(1,3)
Step 2
Step 2.1
Evaluate at x=-7.
Step 2.1.1
Substitute -7 for x.
23√(-7)-1+3
Step 2.1.2
Simplify.
Step 2.1.2.1
Simplify each term.
Step 2.1.2.1.1
Subtract 1 from -7.
23√-8+3
Step 2.1.2.1.2
Rewrite -8 as (-2)3.
23√(-2)3+3
Step 2.1.2.1.3
Pull terms out from under the radical, assuming real numbers.
2⋅-2+3
Step 2.1.2.1.4
Multiply 2 by -2.
-4+3
-4+3
Step 2.1.2.2
Add -4 and 3.
-1
-1
-1
Step 2.2
Evaluate at x=9.
Step 2.2.1
Substitute 9 for x.
23√(9)-1+3
Step 2.2.2
Simplify.
Step 2.2.2.1
Simplify each term.
Step 2.2.2.1.1
Subtract 1 from 9.
23√8+3
Step 2.2.2.1.2
Rewrite 8 as 23.
23√23+3
Step 2.2.2.1.3
Pull terms out from under the radical, assuming real numbers.
2⋅2+3
Step 2.2.2.1.4
Multiply 2 by 2.
4+3
4+3
Step 2.2.2.2
Add 4 and 3.
7
7
7
Step 2.3
List all of the points.
(-7,-1),(9,7)
(-7,-1),(9,7)
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: (9,7)
Absolute Minimum: (-7,-1)
Step 4
