Question

(a) $f(x) = \begin{cases} 4\sqrt{x}, & 0 \le x \le 1\\ 6x - 2, & x > 1. \end{cases}$ (b) $f(x) = \begin{cases} -2x^2 + 4, & 0 \le x \le 1\\ 2x, & x > 1. \end{cases}$

          (a) $f(x) = \begin{cases} 4\sqrt{x}, & 0 \le x \le 1\\ 6x - 2, & x > 1. \end{cases}$
(b) $f(x) = \begin{cases} -2x^2 + 4, & 0 \le x \le 1\\ 2x, & x > 1. \end{cases}$
        
(a) f(x) =  4√(x),     0 ≤ x ≤ 1
 6x - 2,     x > 1.
(b) f(x) =  -2x^2 + 4,     0 ≤ x ≤ 1
 2x,     x > 1.

Added by Joshua P.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Use the definition of a derivative to show that the functions below are not differentiable at x= 1. (a) f(x)={(4sqrt(x),0<=x<=1),(6x-2,x>1):} (b) f(x)={(-2x^(2)+4,0<=x<=1),(2x,x>1):} 4Vx,0< x 1 (a) f(x) = 6x - 2, x > 1. -2x2+4, 0 x 1 2x,x > 1.
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Transcript

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00:02 Okay, we want to show this piecewise defined function f of x is not differentiable at x equals 1.
00:12 For a function to be differentiable, it must be continuous at the point of differentiability.
00:23 So if we can show that f of x is not continuous at 1, then it can't be differentiable at 1.
00:33 Now if we take the limit of f of x as x approaches one from the positive side, as x is approaching one from the positive side, the right side of one, that means x is greater than one.
00:54 We have to use this definition of f of x.
00:58 So as x approaches one from the positive side, f of x will approach two times one or two...
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