Super Easy Derivative that has given me a headache

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Homework Help Overview

The discussion revolves around finding the derivative of the function f(x) = x^(1/3)(x-2), including determining critical numbers, intervals of increase and decrease, local extremes, and concavity. Participants are exploring the application of differentiation techniques, particularly the product and quotient rules.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the first and second derivatives using the product and quotient rules but expresses confusion over their calculations. Some participants question the clarity of the function's notation and suggest that proper formatting is necessary for understanding.

Discussion Status

Participants are actively engaged in clarifying the function's expression and the original poster's attempts at differentiation. There is a recognition of the need for clearer communication, particularly regarding the use of LaTeX for mathematical notation. The discussion is ongoing, with no consensus reached yet.

Contextual Notes

The original poster has indicated a struggle with formatting their question clearly and has expressed a willingness to learn LaTeX if necessary. There is an acknowledgment of potential misunderstandings due to notation issues.

Lena1992
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Super Easy Derivative that has given me a headache!

Homework Statement



So I have to get the derivative for f(x) = x^3/1 *(x-2)
I have to find:
a) Critical Numbers
b) Intervals where it decreases/ increases
c) Local extremes
d) Concave intervals
Then the second derivative of that to get the concave intervals

The Attempt at a Solution



I tried doing the second derivative but I seem to get a wrong answer everytime.I get f(x) = 4x-2 / 3x^2/3 for the first derivative

But I get (12x^2/3) / x^4/3 + 4^x1/3

Can somebody help? I just need to know how to get the second derivative... And no I'm not asking for someone to do my Homework.

Edit: I will post my procedure

f(x)= x^1/3(x-2)
f'(x)= (1/3x^-2/3)(x-2) + (x^1/3)(1) Here I used the product rule (f*g)' = f'*g + f*g'
f'(x)= x-2/3x^2/3 + x^1/3
f'(x)= x-2/3x^2/3 + x^1/3*(3x^2/3)/(3x^2/3) I multiply by (3x^2/3)/(3x^2/3) to make it equal
f'(x)= x-2+3x/ 3x^2/3
f'(x)= 4x-2/3x^2/3

f'(x)=4x-2/3x^2/3
f''(x)= (4)(3x^2/3) - (2^-1/3)(4x-2) / (3x^2/3)^2 Here I use the quotient rule f'(x)= f'*g - g'*f / g^2
f''(x)= 12x^2/3 - (8x^-4/3 - 4x^-1/3) / 9x^4/3
f''(x)= 12x^2/3 / 9x^4/3 - 8x^4/3 + 4x^1/3
f''(x)= 12x^2/3 / x^4/3 + 4x^1/3
 
Last edited:
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Please use LaTeX (or at least parentheses) to correctly format your question. At the moment it is not clear what you mean.

Is x^3/1 *(x-2)

[tex]x^3(x-2)[/tex]

or

[tex]\frac {x^3} {x-2}[/tex]

or something else?
 


I would love to use LaTex but how can I use it to explain myself better?

Edit: I posted my entire procedure I hope it's clear. No it's none of those its the cubic root of x (not x^2). or (x^2/3)

Edit 2: I don't know how to use LaTex but if i NEED to (if the above isn't clear enough) then I will be forced to learn and come back to post it as LaTex.
 
Last edited:


Lena1992 said:
x^3/1 *(x-2)

is different from

f(x)= x^1/3(x-2)

which is

[tex]x^{\frac 1 3}(x-2)[/tex]

Is that it?

Do you see why we have no idea what the question is?
 


[tex]x^{\frac 1 3}(x-2)[/tex]

Yes it is this. I sincerely apologize for my mistakes.

I'm just trying to derive this using the quotient rule please I need someone to help me!

23mtoa8.jpg
 
Last edited:

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