Asked by timmy
f(x)= xsquared -2x
for y= absolute value of f(x) does the derivitive exist at x=0 and explain why.
for y= absolute value of f(x) does the derivitive exist at x=0 and explain why.
Answers
Answered by
MathMate
f(x)=|x²-2x|
=x²-2x (-∞,0]U[2,∞)
=-(x²-2x) (0,2)
f'(x)
=2x-2 (-∞,0)U(2,∞)
=-(2x-2) (0,2)
f'(0-)=-2
f'(0+)=2
Therefore the derivative is not continuous at x=0, and therefore f'(0) does not exist.
See:
http://img683.imageshack.us/img683/654/1285202904.png
=x²-2x (-∞,0]U[2,∞)
=-(x²-2x) (0,2)
f'(x)
=2x-2 (-∞,0)U(2,∞)
=-(2x-2) (0,2)
f'(0-)=-2
f'(0+)=2
Therefore the derivative is not continuous at x=0, and therefore f'(0) does not exist.
See:
http://img683.imageshack.us/img683/654/1285202904.png
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.