derivatives: f(x) = x^3 + 3x^2 / 2(x^2 + x - 6)

[Guest]

I am having a little trouble getting the first & second derivative of an equation to find critical points for concavity and increasing/decreasing intervals. The equation is

f(x)=x^3+3x^2 / 2(x^2+x-6)

I am using d/dx (Fx/Gx)=(Gx)(F'x)-(Fx)(G'x) /(Gx)^2, but i am getting an answer that has a power higher than what i began with,

d/dx= x^4+2x^3-15x^2-36x/x^4+2x^3-11x^2-12x-36 is what i got??????

Please help!

 

[royhaas]

It might be easier if you simply first. Factor the numerator and denominator to end up with \(\displaystyle f(x) = x^2/(2(x-2))\).

This can be simplified further to give \(\displaystyle f(x) = x/2+1 + 2/(x-2)\).

 

[soroban]

Re: derivatives

Hello, blowfish!

\(\displaystyle \:f(x)\:=\:\frac{x^3\,+\,3x^2}{2(x^2\,+\,x\,-\,6)}\)

d/dx = (x^4+2x^3-15x^2-36x)/(x^4+2x^3-11x^2-12x-36) is what i got?
Of course, you're having trouble if you multiply out everything like that!
How do you expect to solve for x with that quartic equation?

Take royhaas' suggestion: reduce the function first.

We have: \(\displaystyle \L\,f(x) \;= \;\frac{x^2(x\,+\,3)}{2(x\,+\,3)(x\,-\,2)} \;= \;\frac{x^2}{2(x\,-\,2)}\)

\(\displaystyle \;\;\)Note: we can cancel because \(\displaystyle x\,=\,-3\) is not in the domain of the function.


Differentiate (quotient rule): \(\displaystyle \L\,f'(x)\;=\;\frac{2(x\,-\,2)\cdot2x\,-\,x^2\cdot2}{4(x\,-\,2)^2} \;= \;\frac{2x^2\,-\,8x}{2(x\,-\,2)^2}\)

\(\displaystyle \;\;\)and we have: \(\displaystyle \L\,f'(x)\;=\;\frac{2(x^2\,-\,x)}{4(x\,-\,2)^2}\;=\;\frac{x^2\,-\,x}{2(x\,-\,2)^2}\)



Differentiate again: \(\displaystyle \L\,f''(x)\;=\;\frac{2(x\,-\,2)^2\cdot(2x\,-\,1)\,-\,(x^2\,-\,x)\cdot4(x\,-\,2)}{4(x\,-\,2)^4}\)

Factor: \(\displaystyle \L\,f''(x)\;= \;\frac{2(x\,-\,2)\,\left[(x\,-\,2)(2x\,-\,1)\,-\,2(x^2\,-\,x)\right]}{4(x\,-\,2)^4}\)

\(\displaystyle \;\;\)and simplify: \(\displaystyle \L\,f''(x)\;= \;\frac{-3x\.+\.2}{2(x\,-\,2)^3}\)


Now you can check for critical points, increasing/decreasing. concavity, etc.

 

[Guest]

Thank you so much for your help. I was making it too difficult. I wasn't even thinking about reducing the equation first. Thanks again!

 

[Guest]

I do have another ? about this one. The more I got to looking at it, i realized I didn't get the same answer. I got the same derivative but the difference came when I factored. if the derivative is (2x^2-8x)/(2x-4)^2, then when it is factored out is should be [2x(x-8)]/[2(x-2)^2]......then when finding the second derivative, it should be

f"(x)=[(2x-4)^2(4x-8)-(2x^2-8x)(2)(2x-4)(2)]/(2x-4)^4, resulting in my final answer of 32/ 2(x-4)^3 (after factoring out (2x-4))......?????????????

 

[skeeter]

if the derivative is (2x^2-8x)/(2x-4)^2, then when it is factored out is should be [2x(x-8)]/[2(x-2)^2] ...

uhh, no.

2x² - 8x = 2x(x - 4)

and

(2x - 4)² = [2(x - 2)]² = 4(x - 2)²

in this case, factoring doesn't help much ... nothing cancels except the 2 in the numerator with a 2 in the denominator ...

x(x - 4)/[2(x - 2)²]

 

[Guest]

You are right. I had a typo. I did have 2x(x-4). All I was trying to say is that is looked incorrect to me. If that factoring was incorrect, then the first and second derivatives would be wrong. Thanks.