# For f(x) = abs(x^3 - 9x), does f'(0) exist

• needingtoknow
In summary: I was able to solve the problem correctly. In summary, to find the derivative at a point, we can use the left and right hand derivatives and compare the results to determine if the derivative exists at that point. It is important to pay attention to the signs of the terms when simplifying the limit expressions.

## Homework Statement

For f(x) = abs(x^3 - 9x), does f'(0) exist?

## The Attempt at a Solution

[/B]
The way I tried to solve this question was to find the right hand and left hand derivative at x = 0.

Right hand derivative
= (lim h--> 0+) f(h) - f(0) / h
= (lim h--> 0+) abs(h^3 - 9h) / h
= (lim h--> 0+) h^2 - 9
= (lim h--> 0+) h^2 - 9 = -9

Left hand derivative
= (lim h--> 0-) f(h) - f(0) / h
= (lim h--> 0-) abs(h^3 - 9h) / h
= (lim h--> 0-) -(h^2 - 9)
= (lim h--> 0-) -h^2 + 9 = 9

However, when I plug in the equation into the graphing calculator, the magnitude is correct, by the positive and negative signs are switched.

needingtoknow said:

## Homework Statement

For f(x) = abs(x^3 - 9x), does f'(0) exist?

## The Attempt at a Solution

[/B]
The way I tried to solve this question was to find the right hand and left hand derivative at x = 0.

Right hand derivative
= (lim h--> 0+) f(h) - f(0) / h
= (lim h--> 0+) abs(h^3 - 9h) / h
= (lim h--> 0+) h^2 - 9

Note that when ##h## is small positive that ##h^3 - 9h < 0## so the last step should have ##9-h^2##.
= (lim h--> 0+) h^2 - 9 = -9

Left hand derivative
= (lim h--> 0-) f(h) - f(0) / h
= (lim h--> 0-) abs(h^3 - 9h) / h
= (lim h--> 0-) -(h^2 - 9)
= (lim h--> 0-) -h^2 + 9 = 9
Similar comment for the second one.

needingtoknow
Thanks a million! This made perfect sense