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

[needingtoknow]

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.

 

[LCKurtz]

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