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

  • #1

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.
 

Answers and Replies

  • #2
LCKurtz
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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.
 
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Likes needingtoknow
  • #3
Thanks a million! This made perfect sense
 

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