Help Deriving and Solving a Function

In summary, the conversation is about deriving a function and finding its derivative using the quotient rule. The correct derivative is found to be (2x^3 - 24x + 32)/(x-2)^6 and the critical points are found to be x=2 and x=-4. A second derivative computation is suggested to determine the nature of the critical point at x=-4.
  • #1
PhysicsinCalifornia
58
0
Please Help Me !

Hello,

I REALLY need help deriving a function

if [tex]f(x) = \frac{-x-2}{(x-2)^3}[/tex]
what would its derivative be?

Here's what I did so far::
Using the quotient rule,
[tex]f'(x) = \frac{(x-2)^3(-1) - (-x-2)(3(x-2)^2(1))}{((x-2)^3)^2)}[/tex]
[tex]= \frac{-(x-2)^3 + (x+2)(3(x^2-4x+4))}{(x-2)^6}[/tex]
[tex]= \frac{-(x^2 -4x+4)(x-2) + (x+2)(3x^2 -12x +12)}{(x-2)^6}[/tex]
[tex]= \frac{-(x^3 -4x^2 + 4x -2x^2 +8x -8) + (3x^3 -12x^2 + 12x +6x^2 -24x +24)}{(x-2)^6}[/tex]
[tex]= \frac{-x^3 + 4x^2 -2x + 2x^2 -8x + 8 + 3x^3 - 12x^2 + 12x + 6x^2 -24x +24}{(x-2)^6}[/tex]
[tex]f'(x) = \frac{2x^3 - 24x + 32}{(x-2)^6}[/tex]

Is this the derivative, and if so, how can I find the critical numbers? (When the derivative equals 0 or DNE(does not exist))
 
Last edited:
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  • #2
You took the derivative correctly, and nice job with the latex, but Ican't check through allthat simplification in my head. To find critical points, set the expression equal to zero and solve. It equals zero whenever the numerator equals zero (and the denominator isnt).
 
  • #3
whozum said:
You took the derivative correctly, and nice job with the latex, but Ican't check through allthat simplification in my head. To find critical points, set the expression equal to zero and solve. It equals zero whenever the numerator equals zero (and the denominator isnt).

I first tried to solve when [tex]2x^3 -24x +32 = 0[/tex]
I thought it was easier to factor the numerator first

Getting: [tex]2(x-2)^2 (x+4) = 0[/tex]
There, much easier.

Now it's clear that x=2 and x=-4

Thanks for your help Whozum :smile:
 
  • #4
You needn't have done all those additions & multiplications,u should have factored [itex] x^{2}-4x+4 [/itex] in the 3-rd line and it would have been simpler.

Daniel.
 
  • #5
PhysicsinCalifornia said:
I first tried to solve when
I thought it was easier to factor the numerator first

Getting:
There, much easier.

Now it's clear that x=2 and x=-4
Warning! Those are not the only critical points. A "critical point" is, as you said, a point where the derivative is 0 (i.e. the numerator is 0) or where the derivative does not exist (i.e. where the denominator is 0).
 
  • #6
The function is not defined in the point [itex] x=2 [/itex],so you could simplify through the monom [itex] (x-2) [/itex] in the final expression of the derivative.

Daniel.
 
  • #7
I got the following

6(x-2)^-4 + 3x(x-2)^-4 - (x-2)^-3

= (3x + 6 - x + 2) (x-2)^-4

= (2x+8)(x-2)^-4

I think my way is easier.
 
  • #8
It looks okay.So the only critical point is [itex] x=-4 [/itex].A second derivative computation will reveal its nature.

Daniel.
 

Related to Help Deriving and Solving a Function

1. How do I derive a function?

To derive a function, you must use the rules of differentiation to find the derivative of the function with respect to the independent variable. These rules include the power rule, product rule, quotient rule, and chain rule. By applying these rules, you can find the slope of the function at any point.

2. What is the purpose of deriving a function?

Deriving a function allows you to find the rate of change of the function at any point. This is useful in many areas of science, such as physics, chemistry, and economics, where understanding how a variable changes over time is important.

3. How do I solve a function?

Solving a function means finding the value(s) of the independent variable that make the function equal to a given value. This can be done by setting the function equal to the given value and solving for the independent variable using algebraic techniques.

4. Why is it important to solve a function?

Solving a function allows you to find specific points on the function, which can be useful in practical applications. For example, in physics, solving a function can help determine the time at which an object reaches a certain position or velocity.

5. What are some common mistakes when deriving and solving a function?

Some common mistakes include forgetting to apply the rules of differentiation correctly, forgetting to simplify the derivative, and making algebraic errors when solving for the independent variable. It is important to check your work carefully to avoid these mistakes.

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