# Thread: Zeroes of a function

1. ## Zeroes of a function

First of all, apologies if this is in the wrong section It's a calculus problem but I'm stuck on the Algebra of it.

I'm supposed to find the critical numbers of f(x)=x^(3/4)-2x^(1/4).
I found the derivative to be f'(x)= (3/4)x^(-1/4)-(1/2)x^(-3/4), but for the life of me I can't remember how to go about finding it's zeroes.
I know one of them is x=0, and the book says another is x=4/9, but how do I get that? Thanks for any and all help in advance.

2. ## Re: Zeroes of a function

$\displaystyle f'(x)= \frac{3}{4}x^{\frac{-1}{4}}-\frac{1}{2}x^{\frac{-3}{4}}$

$\displaystyle 0=\frac{3}{4}x^{\frac{-1}{4}}-\frac{1}{2}x^{\frac{-3}{4}}$

$\displaystyle 0=3x^{\frac{-1}{4}}-2x^{\frac{-3}{4}}$

For a clearer understanding of what we have:

$\displaystyle 0=\frac{3}{\sqrt[4]{x}}-\frac{2}{\sqrt[4]{x^3}}$

$\displaystyle \frac{2}{\sqrt[4]{x^3}}=\frac{3}{\sqrt[4]{x}}$

$\displaystyle 2\sqrt[4]{x}=3\sqrt[4]{x^3}$

Raising both sides to the 4th power:

$\displaystyle 16x=81x^3$

$\displaystyle 81x^3-16x=0$

$\displaystyle x(81x^2-16)=0$

3. ## Re: Zeroes of a function

How did you go from 2/x^(3/4)=3/x^(1/4) to 2x^(1/4)=3x^(3/4)?

4. ## Re: Zeroes of a function

Originally Posted by TheDarkNight93
How did you go from 2/x^(3/4)=3/x^(1/4) to 2x^(1/4)=3x^(3/4)?
I would say he cross-multiplied the fractions ...

if $\displaystyle \frac{a}{b} = \frac{c}{d}$ , then $\displaystyle ad = bc$

5. ## Re: Zeroes of a function

Thanks, thats a lot of help. It's embaressing that I don't remember this (3-4 years ago). Could anyone point me towards a basic algebra review to help refresh my memory on this kind of stuff?