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Problem Set One

Problem Set One. If f(x) = 2|x-1| + (x-1)² then what is the value of f’(0)?. Ellen Dickerson. Finding the derivative of an absolute value. Lets first try and find f’(x), when f(x)=|x| Squaring a number always makes it positive |x| = √ (x)² So f(x) = √ (x)²

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Problem Set One

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  1. Problem Set One If f(x) = 2|x-1| + (x-1)² then what is the value of f’(0)? Ellen Dickerson

  2. Finding the derivative of an absolute value • Lets first try and find f’(x), when f(x)=|x| • Squaring a number always makes it positive • |x| = √ (x)² • So f(x) = √ (x)² • Now we can use the chain rule to find the derivative • Reminder (f∘g)’(x) = f’(g(x))g’(x) • f(x) = (x²) ½ • f’(x) = ½(x²)-½ * g’(x²) • f’(x) = ½(x²) -½ * 2x • f’(x) = ½ (2x)/√ (x)² • f’(x) = x/√ (x)² • f’(x) = x/|x| • f’(|x|) = x/|x|

  3. f(x) = 2|x-1| + (x-1)² • Lets first find the derivative of 2|x-1| • Reminder: f’(|x|) = x/|x| Let u = x-1 f’(u) = 2(u/|u|) f’(u) = 2u/|u| f’(x) = 2(x-1)/|x-1| f’(x) = (2x-2)/|x-1| f’(2|x-1|) = (2x-2)/|x-1| • Now we will find the derivative of (x-1)2 • Reminder (f∘g)’(x) = f’(g(x))g’(x) f’(x) f(x) = (x-1)2 f’(x) = 2(x-1) * 1 f’(x) = 2x-2 f’((x-1)2) = 2x-2

  4. f’(2|x-1|) = (2x-2)/|x-1| • f’((x-1)2) = 2x-2 • So if f(x) = 2|x-1| + (x-1)² • f’(x) = (2x-2)/|x-1| + 2x-2 • Now to find f’(0) we need to plug in a 0 to x • f’(0) = (2(0)-2)/|0-1| + 2(0)-2 • f’(0) = (-2)/(-1) + (-2) • f’(0) = 2+(-2) • f’(0) = 0

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