1. ## Finding the Derivative

I am stuck on this one problem that I have for homework. The question says to find the critical numbers for f(x)= |3x-4|. I do not know how to take the derivative of an absolute value function. Any help would be greatly appreciated!!! Thank you in advance.

2. Originally Posted by jmhogart
I am stuck on this one problem that I have for homework. The question says to find the critical numbers for f(x)= |3x-4|. I do not know how to take the derivative of an absolute value function. Any help would be greatly appreciated!!! Thank you in advance.
I'd imagine you take it in the normal way but bearing in mind that the range is not negative

3. So would that be [3(3x-4) / |3x-4|] or just 3?

4. Well, I should think you would have to divide the derivative into different functions derived directly from d(3x-4)/dx, y' being multiplied by negative one for domains where y < 0, remaining the same where y > 0, and being nonexistent whenever y = 0.

 Actually, I think d|u|/dx still exists at u = 0 so long as du/dx = 0. (that is, if the function has a local min or max at an x-intercept)

5. Originally Posted by jmhogart
I am stuck on this one problem that I have for homework. The question says to find the critical numbers for f(x)= |3x-4|. I do not know how to take the derivative of an absolute value function. Any help would be greatly appreciated!!! Thank you in advance.
$\displaystyle f(x) = 3x - 4$ when $\displaystyle 3x - 4 > 0 \Rightarrow x > \frac{4}{3}$.

$\displaystyle f(x) = -(3x - 4) = 4 - 3x$ when $\displaystyle 3x - 4 \leq 0 \Rightarrow x \leq \frac{4}{3}$.

The derivative is not defined at the salient point of f(x), ie. at (4/3, 0).