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SOLUTION If x > 0, then |x| = and we can choose h small enough that x + h > 0 and hence |x + h| = . Therefore, for x > 0 we have |x + h\ – |x| f'(x) = lim h-0 (x + h) – x = lim h-0 h = lim h-0 = lim h-0 and so f is differentiable for any x > Similarly, for x < we have |x| = and h can be chosen small enough that x + h < and so Ix + h| = . Therefore for x < 0, |x + h| - |x| f'(x) = lim -(x + h) – (-x) = lim h-0 h = lim h-0. h = lim h-0 and so fis differentiable for any x < 0. For x = 0 we have to investigate f(0 + h) - f(0) f'(x) = lim h-0 h 10 + h| - 10| lim (if it exicte)

SOLUTION If x > 0, then |x| = and we can choose h small enough that x + h > 0 and hence |x + h| = . Therefore, for x > 0 we have |x + h\ – |x| f'(x) = lim h-0 (x + h) – x = lim h-0 h = lim h-0 = lim h-0 and so f is differentiable for any x > Similarly, for x < we have |x| = and h can be chosen small enough that x + h < and so Ix + h| = . Therefore for x < 0, |x + h| - |x| f'(x) = lim -(x + h) – (-x) = lim h-0 h = lim h-0. h = lim h-0 and so fis differentiable for any x < 0. For x = 0 we have to investigate f(0 + h) - f(0) f'(x) = lim h-0 h 10 + h| - 10| lim (if it exicte)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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SOLUTION If x > 0, then |x| and we can choose h small enough that x + h > 0 and hence |x + h| = Therefore, for x > 0 we have |x + h| - |x| f'(x) lim h → 0 %D h (х + h) - lim h → 0 lim h → 0 h lim h → 0 and so f is differentiable for any x > Similarly, for x < we have |x| and h can be chosen small enough that x + h < and so |x + h| = Therefore for x < 0, |x + h[ – \x] f'(x) lim h → 0 %D h -(x + h) – (-x) lim h → 0 lim h → 0 h lim h → 0 and so f is differentiable for any x < 0. For x = 0 we have to investigate f(0 + h) – f(0) - f'(x) = lim h → 0 h |0 + h| – 10| lim (if it exists). h → 0 II
Transcribed Image Text:SOLUTION If x > 0, then |x| and we can choose h small enough that x + h > 0 and hence |x + h| = Therefore, for x > 0 we have |x + h| - |x| f'(x) lim h → 0 %D h (х + h) - lim h → 0 lim h → 0 h lim h → 0 and so f is differentiable for any x > Similarly, for x < we have |x| and h can be chosen small enough that x + h < and so |x + h| = Therefore for x < 0, |x + h[ – \x] f'(x) lim h → 0 %D h -(x + h) – (-x) lim h → 0 lim h → 0 h lim h → 0 and so f is differentiable for any x < 0. For x = 0 we have to investigate f(0 + h) – f(0) - f'(x) = lim h → 0 h |0 + h| – 10| lim (if it exists). h → 0 II
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