# Math Help - few calc/alg questions

1. ## few calc/alg questions

1) if you have 4x^(-2), and you wanted to make the power positive, would you bring the coefficient to the bottom?

e.g. 1/(4x^2) -OR- 4/(x^2)

2) find the derivative of (sqrt(x)-1)/(sqrt(x)+1)
I understand the quotient rule and syntax, it just doesn't work out to the answer that I've been given... PROBABLY a simplification error... here's what I did:

{(x½+1)(1/2x-½)-(x½-1)(1/2x-½)}/(x½+1)^2

(1/2)+(1/2x-½)-(1/2)+(1/2x-½)

x-½/(x½+1)^2

what I get:

1/(x^2(x½+1)^2)

3) why is it that when you look at the cuberoot function (x^(1/3)) from left to right, you see its slope is always positive (albeit undefined at x=0), yet the derivative (1/3x^(-1/3)) shows that its slope is negative from a domain of negative infinity to 0? should it not be positive?

thanks.

2. 1) The coefficient stays on top to make $4/(x^2)$

The rest I don't know

3. thank you, anyone know the others?

4. Hello, squidplant!

1) if you have $4x^{-2}$, and you wanted to make the power positive,
would you bring the coefficient to the bottom? . . . No

This is correct: $\frac{4}{x^2}$

2) Find the derivative of: $y \:= \:\frac{\sqrt{x}-1}{\sqrt{x}+1}$

I got: . $y' \:=\:\frac{x^{-\frac{1}{2}}}{\left(x^{\frac{1}{2}}+1\right)^2}$ . . . Correrct!

And that becomes: . $y' \:=\:\frac{1}{x^{\frac{1}{2}}\left(x^{\frac{1}{2}} - 1\right)^2}$
You had it right . . . you moved the $x^{-\frac{1}{2}}$ incorrectly.

3) Why is it that when you look at the cuberoot function $x^{\frac{1}{3}}$ from left to right,
you see its slope is always positive (albeit undefined at x=0),
yet the derivative $\frac{1}{3}x^{-\frac{1}{3}}$ shows that its slope is negative
from a domain of negative infinity to 0? .Should it not be positive?

Your derivative is incorrect: . $y' \:=\:\frac{1}{3}x^{-\frac{2}{3}} \:=\:\frac{1}{3x^{\frac{2}{3}}} \:=\:\frac{1}{3\left(\sqrt[3]{x}\right)^2}$

So the derivative is positive for all $x \neq 0.$

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A simple error, but do not be embarrassed.

You are Thinking about derivatives, slopes, signs, increasing/decreasing, etc.

Evidently you have a thorough understanding of these concepts ... impressive!

And you thought you found a contradiction . . . good Thinking!

5. Thank you very much Sorobon, it's good to know that mathematics still remains my haven of consistency and pure pragmitism with all the things occuring in my life; I owe math a cure for cancer. Take care~

(p.s. i'm in my first year of AP Calculus at my high school, junior year. we're learning integrals in a few weeks! ^_^)