Let f:R→Rf:\mathbb{R} o \mathbb{R} be the function f(x)=√x−1f(x)=\sqrt{x-1} for x≥1x\geq 1 and f(x)=|x|−1f(x) = |x| - 1 for x

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Let f:R→Rf:\mathbb{R}	o \mathbb{R} be the function f(x)=√x−1f(x)=\sqrt{x-1} for x≥1x\geq 1 and f(x)=|x|−1f(x) = |x| - 1 for x<1x< 1. Determine all values of xx where ff is differentiable.

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We start by restating the problem in our own words: f is defined piecewise as f(x) = sqrt{x-1} for x ≥ 1, and f(x) = |x| - 1 for x < 1. We need to find all x where f is differentiable. First, consider the region x > 1. Here f(x) = sqrt{x-1}, which is a composition of smooth functions on (1, ∞). Its derivative is f'(x) = 1/(2 sqrt{x-1}), which exists for x > 1. Therefore, f is differentiable on (1, ∞). Next, look at the region x < 1. Here f(x) = |x| - 1. The function |x| is differentiable for all x ≠ 0, with derivative -1 for x < 0 and +1 for x > 0. Thus, for x < 1 we have: - If x < 0, f(x) = -x - 1 with derivative -1, so differentiable on (-∞, 0). - If 0 < x < 1, f(x) = x - 1 with derivative 1, so differentiable on (0, 1). - At x = 0, there is a cusp in |x|, since the left-hand derivative is -1 and the right-hand derivative is +1; hence f is not differentiable at x = 0. Now check the junction at x = 1. We must compare left and right derivatives and function values: - From the left (x < 1), as x approaches 1 from below, we are in the region 0 ≤ x < 1 where f(x) = x - 1, whose derivative is 1. So the left-hand derivative at x = 1 is 1. - From the right (x > 1), f(x) = sqrt{x-1}, whose derivative is 1/(2 sqrt{x-1}). As x → 1+, this derivative tends to +∞, i.e., it does not exist as a finite number. - Since the right-hand derivative does not exist (and in particular is not equal to the left-hand derivative), f is not differentiable at x = 1. Therefore, the set of differentiable points is where both sides behave nicely: on (-∞, 0) it’s differentiable with derivative -1, on (0, 1) it’s differentiable with derivative 1, and on (1, ∞) it’s differentiable with derivative 1/(2 sqrt{x-1}). We must exclude the points 0 and 1 where differentiability fails. Putting these together, f is differentiable for all real x except x = 0 and x = 1. In set notation: R \ {0, 1}. This matches option a: R \ {0, 1}.

ENG1090 - MUM S1 2025 Lecture quiz 7

Let f:R→Rf:\mathbb{R}\to \mathbb{R} be the function f(x)=√x−1f(x)=\sqrt{x-1} for x≥1x\geq 1 and f(x)=|x|−1f(x) = |x| - 1 for x<1x< 1. Determine all values of xx where ff is differentiable.

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