I am asked to simplify this:
View attachment 12199
Here is what I got...and this is only the latest in a series:
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View attachment 12200
I am full of doubt about this because...why would they have that (4-x^2) in both numerator and denominator if it wasn't meant to disappear somehow? Anyway, this is my best shot so far.
Yes, it does look nasty, but tkhunny is correct that this sort of ugliness arises in calculus. Technically, the expression above is the derivative of
[MATH]\dfrac{(6x + 1)^{1/3}}{4 - x^2}.[/MATH]
The derivative of a function is the function's "slope."
What I would do
initially is to break the fraction into two expressions that will be a bit easier to work with, then simplify, then get common denominators, simplify again, and add. Trying to work on that mess in one fell swoop will drive you bonkers
quickly.
[MATH]\dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6) - (6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = [/MATH]
[MATH]\dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6)}{(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = [/MATH]
[MATH]\dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = [/MATH]
[MATH]\dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{2/3} * (6x + 1)^{1/3}(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} = [/MATH]
[MATH]\dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} = [/MATH]
[MATH]\dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{-\ 12x^2 - 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = [/MATH]
[MATH]\dfrac{8 - 2x^2 + 12x^2 + 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = \dfrac{10x^2 + 2x + 8}{(6x + 1)^{2/3}(4 - x^2)^2}[/MATH]
I shall test it using x = 0.
[MATH]\dfrac{(4 - 0^2)\left( \frac{1}{3} \right )(6 * 0 + 1)^{-2/3}(6) - (6 * 0 + 1)^{1/3}(-\ 2 * 0)}{1^{2/3} * (4 - 0^2)^2} = \dfrac{2(4)(1)^{-2/3}}{1 * 4^2} = \dfrac{1}{2}.[/MATH]
[MATH]\dfrac{10 * 0^2 + 2 * 0 + 8}{(6 * 0 + 1)^{2/3}(4 - 0^2)^2} = \dfrac{8}{1 * 16} = \dfrac{1}{2}.[/MATH]
Your algebra looks good to me. What was your test number on the calculator?