Let $\operatorname{sgn} u$ be 1, 0 or $-1$ as $u$ is positive, zero or negative. At how many points is $f(x) = \operatorname{sgn}\left((x - 1)^{2}(x - 4)\right)$ not continuous?

0

Correct answerYour answer

1

Correct answerYour answer

2

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3

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$f$ can break only where $u = (x - 1)^{2}(x - 4)$ is zero: at $x = 1$ and $x = 4$ .
At 4, $u$ changes sign, so $f$ jumps from $-1$ to 1. At 1, $u < 0$ on both sides, so $f \to -1$ , but $f(1) = 0$ : a break as well. So 2 points.

Worth noting

1 counts only the sign change at 4; at the double root 1 the sign does not change, but the value 0 still differs from the limit $-1$ .

Ten questions from Continuity & Differentiability, at your own pace.

Let sgn⁡u\operatorname{sgn} usgnu be 1, 0 or −1-1−1 as uuu is positive, zero or negative. At how many points is f(x)=sgn⁡((x−1)2(x−4))f(x) = \operatorname{sgn}\left((x - 1)^{2}(x - 4)\right)f(x)=sgn((x−1)2(x−4)) not continuous?