Approximating Relative Minima or Maxima

[nycmathguy]

Use a graphing utility to approximate (to two decimal places) any relative minima or maxima of the function.

52. f(x) = x^3 − 3x^2 − x + 1

54. g(x) = x•sqrt{4 - x}

A. Define relative minima
B. Define relative maxima
C. Provide steps for doing 52 and 54.

This is it for today.

Thanks

 

[MathLover1]

52. f(x) = x^3 − 3x^2 − x + 1

If f ' (x )>0 to the left of x=c and f ' (x )<0 to the right of x=c then x=c is a local maximum.
If f ' (x )<0 to the left of x=c and f ' (x )>0 to the right of x=c then x=c is a local minimum.
If f ' (x ) is the same sign on both sides of x=c then x=c is neither a local maximum nor a local minimum.

so, f ' (x ) will be
f'(x) = 3x^2 − 2*3x − 1
f'(x) = 3x^2 − 6x − 1

equal to zero and solve for x

3x^2 − 6x − 1=0............using quadratic formula you will get

x = 1 + 2/sqrt(3)
or
x = 1 - 2/sqrt(3)

so f ' (x )>0: if x <1 - 2/sqrt(3) or x> 1 + 2/sqrt(3)
f ' (x )<0: if 1 - 2/sqrt(3)<x<1 + 2/sqrt(3)

substitute x in original equation to find the y coordinates of the max and min

f(x) = (1 + 2/sqrt(3))^3 − 3(1 + 2/sqrt(3))^2 − (1 + 2/sqrt(3)) + 1
f(x) = 20/(3sqrt(3)) - 4 sqrt(3) -2

f(x) = (1 - 2/sqrt(3))^3 − 3(1 - 2/sqrt(3))^2 − (1 - 2/sqrt(3)) + 1
f(x) = - 20/(3 sqrt(3)) + 4 sqrt(3)-2

Maximum: ( 1-2sqrt(3)/3, - 20/(3 sqrt(3)) + 4 sqrt(3)-2)
Minimum 1+2sqrt(3)/3, 20/(3sqrt(3)) - 4 sqrt(3)-2)




54. g(x) = x*sqrt(4 - x)

to find g'(x) apply the product rule

g'(x) = (d/dx)(x)(sqrt(4 - x))+ (d/dx)(sqrt(4 - x)*x

g'(x) = 1*(sqrt(4 - x))+ (-1/(2sqrt(4 - x))*x

g'(x) = sqrt(4 - x)-x/(2sqrt(4 - x)) ....use common denominator rule, and you get

g'(x) = (8 - 3x)/(2sqrt(4 - x))
equal to zero and solve for x

(8 - 3x)/(2sqrt(4 - x))=0 will be zero only if
(8 - 3x)=0
8=3x
x=8/3
2sqrt(4 - x)=0 if
sqrt(4 - x)=0 if 4 - x=0=> x=4

substitute x in original equation to find the y coordinates of the max and min

g(x) = (8/3)*sqrt(4 - 8/3)
g(x) = (8/3)*sqrt(4/3)
g(x) = (8/3)*sqrt(4)/sqrt(3)
g(x) = (8/3)*2/sqrt(3)
g(x) = (8*2)/(3*sqrt(3))
g(x) = 16/(3*sqrt(3))

g(x) = 4*sqrt(4 - 4)
g(x) = 4*sqrt(0)
g(x) = 4*0
g(x) =0

so,
Maximum: ( 8/3, 16/(3*sqrt(3)))
Minimum: (4, 0 )

 

[nycmathguy]

52. f(x) = x^3 − 3x^2 − x + 1

If f ' (x )>0 to the left of x=c and f ' (x )<0 to the right of x=c then x=c is a local maximum.
If f ' (x )<0 to the left of x=c and f ' (x )>0 to the right of x=c then x=c is a local minimum.
If f ' (x ) is the same sign on both sides of x=c then x=c is neither a local maximum nor a local minimum.

so, f ' (x ) will be
f'(x) = 3x^2 − 2*3x − 1
f'(x) = 3x^2 − 6x − 1

equal to zero and solve for x

3x^2 − 6x − 1=0............using quadratic formula you will get

x = 1 + 2/sqrt(3)
or
x = 1 - 2/sqrt(3)

so f ' (x )>0: if x <1 - 2/sqrt(3) or x> 1 + 2/sqrt(3)
f ' (x )<0: if 1 - 2/sqrt(3)<x<1 + 2/sqrt(3)

substitute x in original equation to find the y coordinates of the max and min

f(x) = (1 + 2/sqrt(3))^3 − 3(1 + 2/sqrt(3))^2 − (1 + 2/sqrt(3)) + 1
f(x) = 20/(3sqrt(3)) - 4 sqrt(3) -2

f(x) = (1 - 2/sqrt(3))^3 − 3(1 - 2/sqrt(3))^2 − (1 - 2/sqrt(3)) + 1
f(x) = - 20/(3 sqrt(3)) + 4 sqrt(3)-2

Maximum: ( 1-2sqrt(3)/3, - 20/(3 sqrt(3)) + 4 sqrt(3)-2)
Minimum 1+2sqrt(3)/3, 20/(3sqrt(3)) - 4 sqrt(3)-2)




54. g(x) = x*sqrt(4 - x)

to find g'(x) apply the product rule

g'(x) = (d/dx)(x)(sqrt(4 - x))+ (d/dx)(sqrt(4 - x)*x

g'(x) = 1*(sqrt(4 - x))+ (-1/(2sqrt(4 - x))*x

g'(x) = sqrt(4 - x)-x/(2sqrt(4 - x)) ....use common denominator rule, and you get

g'(x) = (8 - 3x)/(2sqrt(4 - x))
equal to zero and solve for x

(8 - 3x)/(2sqrt(4 - x))=0 will be zero only if
(8 - 3x)=0
8=3x
x=8/3
2sqrt(4 - x)=0 if
sqrt(4 - x)=0 if 4 - x=0=> x=4

substitute x in original equation to find the y coordinates of the max and min

g(x) = (8/3)*sqrt(4 - 8/3)
g(x) = (8/3)*sqrt(4/3)
g(x) = (8/3)*sqrt(4)/sqrt(3)
g(x) = (8/3)*2/sqrt(3)
g(x) = (8*2)/(3*sqrt(3))
g(x) = 16/(3*sqrt(3))

g(x) = 4*sqrt(4 - 4)
g(x) = 4*sqrt(0)
g(x) = 4*0
g(x) =0

so,
Maximum: ( 8/3, 16/(3*sqrt(3)))
Minimum: (4, 0 )

1. Thank you for your detailed reply but please do not show the entire solution unless I ask for it. The purpose here is for me to do the math, right? If I say, Mira, this is way over my head. Can you provide the solution? Of course, you can then do all the math work. Understand?

2. If you are going to show an entire solution, leave at least one question for me to do. Doing the math is the only way I will ever get ready for Calculus 1 in the near future.

3. You did not use the "Math For Dummies" method. Your definition of minima and maxima went right over my head, honestly.

4. Why bring calculus into this reply? I am not mathematically ready for derivatives.

5. Are you interested in the MathMagic Lite app? It's free. It's awesome. It's easy to use. I don't understand all the features but I do know how to use the app to write math symbols.

Here is a sample:

 

[nycmathguy]

52. f(x) = x^3 − 3x^2 − x + 1

If f ' (x )>0 to the left of x=c and f ' (x )<0 to the right of x=c then x=c is a local maximum.
If f ' (x )<0 to the left of x=c and f ' (x )>0 to the right of x=c then x=c is a local minimum.
If f ' (x ) is the same sign on both sides of x=c then x=c is neither a local maximum nor a local minimum.

so, f ' (x ) will be
f'(x) = 3x^2 − 2*3x − 1
f'(x) = 3x^2 − 6x − 1

equal to zero and solve for x

3x^2 − 6x − 1=0............using quadratic formula you will get

x = 1 + 2/sqrt(3)
or
x = 1 - 2/sqrt(3)

so f ' (x )>0: if x <1 - 2/sqrt(3) or x> 1 + 2/sqrt(3)
f ' (x )<0: if 1 - 2/sqrt(3)<x<1 + 2/sqrt(3)

substitute x in original equation to find the y coordinates of the max and min

f(x) = (1 + 2/sqrt(3))^3 − 3(1 + 2/sqrt(3))^2 − (1 + 2/sqrt(3)) + 1
f(x) = 20/(3sqrt(3)) - 4 sqrt(3) -2

f(x) = (1 - 2/sqrt(3))^3 − 3(1 - 2/sqrt(3))^2 − (1 - 2/sqrt(3)) + 1
f(x) = - 20/(3 sqrt(3)) + 4 sqrt(3)-2

Maximum: ( 1-2sqrt(3)/3, - 20/(3 sqrt(3)) + 4 sqrt(3)-2)
Minimum 1+2sqrt(3)/3, 20/(3sqrt(3)) - 4 sqrt(3)-2)




54. g(x) = x*sqrt(4 - x)

to find g'(x) apply the product rule

g'(x) = (d/dx)(x)(sqrt(4 - x))+ (d/dx)(sqrt(4 - x)*x

g'(x) = 1*(sqrt(4 - x))+ (-1/(2sqrt(4 - x))*x

g'(x) = sqrt(4 - x)-x/(2sqrt(4 - x)) ....use common denominator rule, and you get

g'(x) = (8 - 3x)/(2sqrt(4 - x))
equal to zero and solve for x

(8 - 3x)/(2sqrt(4 - x))=0 will be zero only if
(8 - 3x)=0
8=3x
x=8/3
2sqrt(4 - x)=0 if
sqrt(4 - x)=0 if 4 - x=0=> x=4

substitute x in original equation to find the y coordinates of the max and min

g(x) = (8/3)*sqrt(4 - 8/3)
g(x) = (8/3)*sqrt(4/3)
g(x) = (8/3)*sqrt(4)/sqrt(3)
g(x) = (8/3)*2/sqrt(3)
g(x) = (8*2)/(3*sqrt(3))
g(x) = 16/(3*sqrt(3))

g(x) = 4*sqrt(4 - 4)
g(x) = 4*sqrt(0)
g(x) = 4*0
g(x) =0

so,
Maximum: ( 8/3, 16/(3*sqrt(3)))
Minimum: (4, 0 )

Can you do this without calculus? The instructions are to use a graphing utility.

 

[MathLover1]

Can you do this without calculus? The instructions are to use a graphing utility.

here is the link, place mouse on min and max and it shows you coordinates
https://www.symbolab.com/solver/step-by-step/f\left(x\right) = x^{3} − 3x^{2} − x + 1

 

[nycmathguy]

here is the link, place mouse on min and max and it shows you coordinates
https://www.symbolab.com/solver/step-by-step/f\left(x\right) = x^{3} − 3x^{2} − x + 1

What mouse? I don't have a computer or laptop. Believe me, all my threads (with the exception of attachments) are typed using my cell phone keyboard.

 

[MathLover1]

when you scroll down to see the graph, you will see black dots on max and min, tap that dots to see coordinates

 

[nycmathguy]

when you scroll down to see the graph, you will see black dots on max and min, tap that dots to see coordinates

Can you provide the steps for me to try other problems?

What if the function is h(x) = sqrt{x} + 1?

 

[MathLover1]

1. Thank you for your detailed reply but please do not show the entire solution unless I ask for it. The purpose here is for me to do the math, right? If I say, Mira, this is way over my head. Can you provide the solution? Of course, you can then do all the math work. Understand?

2. If you are going to show an entire solution, leave at least one question for me to do. Doing the math is the only way I will ever get ready for Calculus 1 in the near future.

3. You did not use the "Math For Dummies" method. Your definition of minima and maxima went right over my head, honestly.

4. Why bring calculus into this reply? I am not mathematically ready for derivatives.

5. Are you interested in the MathMagic Lite app? It's free. It's awesome. It's easy to use. I don't understand all the features but I do know how to use the app to write math symbols.

Here is a sample:

View attachment 67

further simplified 6sqrt(8) =6sqrt(2^2*2)=6*2sqrt(2)=12sqrt(2)

 

[nycmathguy]

further simplified 6sqrt(8) =6sqrt(2^2*2)=6*2sqrt(2)=12sqrt(2)

You did not have to further simplify 6•sqrt{8}. I was simply trying to convince you to use the free MathMagic Lite app. Check out my new thread
Precalculus: A Pause In the Road. Tell me what you think. Once again, I can't thank you enough for your help and guidance along the way.