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EXAMPLE 5 Where is the function f(x) = |x-1| differentiable?

EXAMPLE 5 Where is the function f(x) = |x-1| differentiable?

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.5: The Kernel And Range Of A Linear Transformation
Problem 30EQ
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Q Type here to search A y 1- Z Video Example W S X 3 EXAMPLE 5 C SOLUTION x+h-1< F4 Where is the function f(x) = |x-1| differentiable? If x 1, then | x - 1 | = $ 4 f'(x) = 0 lim O BI and we can choose h small enough that x + h - 1 > 0 and hence | x + h- . therefore, for x > 1 we have lim h-0 lim =h→0 and so f is differentiable for any x > Similarly, for x < 1 we have |x-1| = F5 |x+h-1|-|x-1| % 5 h h h h R T and so I x + h-1 | = F6 |x+h-1|-|x-1| F7 n 7 and h can be chosen small enough that FB Therefore for x < 1, U 4 8 68°F Partly sunny D F G H J1 K ²
Transcribed Image Text:Q Type here to search A y 1- Z Video Example W S X 3 EXAMPLE 5 C SOLUTION x+h-1< F4 Where is the function f(x) = |x-1| differentiable? If x 1, then | x - 1 | = $ 4 f'(x) = 0 lim O BI and we can choose h small enough that x + h - 1 > 0 and hence | x + h- . therefore, for x > 1 we have lim h-0 lim =h→0 and so f is differentiable for any x > Similarly, for x < 1 we have |x-1| = F5 |x+h-1|-|x-1| % 5 h h h h R T and so I x + h-1 | = F6 |x+h-1|-|x-1| F7 n 7 and h can be chosen small enough that FB Therefore for x < 1, U 4 8 68°F Partly sunny D F G H J1 K ²
F3 Similarly, for x < 1 we have |x-1|=| x + h-1< lim f'(x) = h→0 = F4 lim h→0 1x+h-1|-|x-1| lim h→0 lim h→0+ and so | x + h-1 | = lim f(1) = h 01 O Et and so f is differentiable for any x < 1. For x = 1 we have to investigate f(1+h)-f(1) lim h→0 F5 % h h -h - h Let's compute the left and right hand limits separately: 10+ h|-|0| h 101-101 h h 10+hI-101 F6 if it exists. Th| = lim = lim = h→0+ h |h| h h→0+ h h F7 & and h can be chosen small enough that lim 1 = h→0+ F8 . Therefore for x < 1, * O F9 68°F Partly sunny F10 F11 A
Transcribed Image Text:F3 Similarly, for x < 1 we have |x-1|=| x + h-1< lim f'(x) = h→0 = F4 lim h→0 1x+h-1|-|x-1| lim h→0 lim h→0+ and so | x + h-1 | = lim f(1) = h 01 O Et and so f is differentiable for any x < 1. For x = 1 we have to investigate f(1+h)-f(1) lim h→0 F5 % h h -h - h Let's compute the left and right hand limits separately: 10+ h|-|0| h 101-101 h h 10+hI-101 F6 if it exists. Th| = lim = lim = h→0+ h |h| h h→0+ h h F7 & and h can be chosen small enough that lim 1 = h→0+ F8 . Therefore for x < 1, * O F9 68°F Partly sunny F10 F11 A
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