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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 60ae
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 60ae

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# Solved: One-sided derivatives The left-hand and right-hand

ISBN: 9780321570567 2

## Solution for problem 60AE Chapter 3.1

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition

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Problem 60AE

One-sided derivatives? ?The left-hand and right-hand derivatives of a function at a point a are given by ? f(a+h)?f(a) f+(a) = lim + h h?0 and f (a) = lim f(a+h)?f(a) ? h?0 + h provided these limits exist. The derivative f??(?a?) ?exists if and only if ?. f+(a) = f (?) a.? ?Sketch the following functions. ? ? b.? ?Compute f (+)and f (a)at?the given point a. c.? ?Is f continuous at a? Is f differentiable at a? ;a=1

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SOLUTION Given and a=1 STEP 1 (a). Sketch the following functions STEP 2 (b). Compute f +a)and f ()at the given point a. The left and the right derivatives of the function are f(a+h)f(a) f(a+h)f(a) f +a) = lim + h And f(a) = lim + h h0 h0 Provided a=1. Therefore 2 2 2 (4(1+h) )(41 ) (4(1+h +2h)4+1 h 2h f +1) = lim + h = lim + h = lim h = lim h 2 = 2 h0 h0 h0 h0 Thus we got f (1) = 2 + 2(1+h)+1(2+1) 2+2h+121 2h Then, f (1 = lim h = lim h = lim h = 2 h0 h0 h0 Thus we got f (1) = 2 STEP 3 (c) Is f continuous at a Is f differentiable at a Now we need to verify whether the function is continuous at a=2 For this we have to show lim f(x) + lim f(x) x1 x1 lim f(x) = lim 4 x = lim 4 1 = 3 + + + x1 x1 x1 lim f(x) = lim 2x + 1 = 2 + 1 = 3 x1 x1 Thus the condition is satisfied. Thus the function f(x) is continuous at a=1. It is already known that the derivative f(a) exists if and only if . f+(a) = f () Since ,in this case f (a) + / f(a) the derivative does not exist. Therefore the given function f(x)is not differentiable at a=1.

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