Please help, the absolute value throws me off.

$\displaystyle g(t) = |3t-4|$

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- May 18th 2012, 12:47 AMrabert1find critical numbers of function with absolute value
Please help, the absolute value throws me off.

$\displaystyle g(t) = |3t-4|$ - May 18th 2012, 12:54 AMGokuRe: find critical numbers of function with absolute value
Draw the graph, then find the "V" point or cusp and that will be your critical point.

- May 18th 2012, 01:06 AMrabert1Re: find critical numbers of function with absolute value
Can you kindly show how to do this with calculus\algebra

- May 18th 2012, 01:19 AMGokuRe: find critical numbers of function with absolute value
- May 18th 2012, 01:22 AMrabert1Re: find critical numbers of function with absolute value
nice find. thanks

- May 18th 2012, 09:04 AMHallsofIvyRe: find critical numbers of function with absolute value
You start with

**definitions**! A "critical point" of a function of one variable is a point where either the derivative is 0 or the derivative does not exist.

The |x| is x if $\displaystyle x\ge 0$ and -x if $\displaystyle x< 0$.

In particular |3t- 4| is 3t- 4 if $\displaystyle 3t- 4\ge 0$ or $\displaystyle t\ge 4/3$, -(3t- 4)= 4- 3t for [tex]t< 4/3[/tex[. So for [tex]t> 4/3[tex], the function is 3t- 4 and has derivative 3. For $\displaystyle x< 3/4$, the function is 4- 3t which has derivative -3. Those are never 0 so the only possible critical point would be at x= 4/3. We need to look at that point separately.

To determine the derivative**at**x= 4/3 we can do either of two things:

1) Use the basic definition. The derivative of f(x) at x= 4/3 is $\displaystyle \lim_{h\to 0}\frac{f((4/3)+h)- f(4/3)}{h}$. f(4/3)= 0, of course, but because of the change of formula on both sides of x= 4/3, we need to look at the "one sided" derivatives. If h> 0 then (4/3)+ h> 4/3, f((4/3)+ h)= 3((4/3)+ h)- 4= 3h so the difference quotient is $\displaystyle \frac{3h}{h}= 3$ and the limit is 3. If h< 0 then (4/3)+ h< 4/3, f((4/3)+h)= 4- 3((4/3)+ h)= -3h so the difference quotient is $\displaystyle \frac{-3h}{h}= -3$ and the limit is -3. Since those two one sided limits are not the same, the limit itself does not exist and the function is not differentiable at x= 4/3.

2) Use a more "sophisticated" theorem. While the derivative of a differentiable function is not itself necessarily continuous, one can use the mean value theorem to show that the "intermediate value theorem" does, in fact, apply. From that, it follows that**if**the derivative at a specific point exists, the two limits $\displaystyle \lim_{x\to a^-}f'(x)$ and $\displaystyle \lim_{x\to a^+}f'(x)$ must both be equal to the derivative at that point. Here, of course, those two limits are -3 and 3 and are not the same so the function is not differentiable at x= 4/3.