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Consider the following function

$$ f(x) = \left\{ \begin{array}{ll} \left|x\right| & \quad x <2\\ \lfloor x\rfloor & \quad x \ge 2 \end{array} \right. $$

where $\lfloor\rfloor$ denotes the greatest integer function. For $x \in(-1,4)$ I need to analyze which options are true ?

(a) $f$ has a local maxima

(b) $f$ has a local minima

(c) $f$ is continuous

(d) $f$ is not differentiable

So, option (c) is false as the greatest integer is discontinuous at integral points.

option (d) is correct as $|x|$ is non differentiable at $x=0$ also , $\lfloor x\rfloor$ is non differentiable at integral points.

(b) Is correct because $f(0-) > f(0)$ and $f(0) < f(0+)$ so $f$ has a local minima at $x=0$

Option (a) also seems to be correct because

for $x \ge 2$ $f = \lfloor x\rfloor$ and

$f(3-) = 2$ while $f(3) = f(3+) =3$ so $f(3-) < f(3)$ and $f(3) \ge f(3+)$ so $x =3$ is one such point of local maxima.

So, according to me options (a) (b) and (d) are correct.

However, my book says correct choices are (b) and (d)

Can anyone please explain why is option (a) incorrect ? and what is wrong with my solution ?

Thank you.

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    $\begingroup$ Your answer is right. c) is false and a), b) d) are all true. $\endgroup$ – Kavi Rama Murthy Dec 6 '19 at 8:12
  • $\begingroup$ At $x=3$ no peak is formed. $\endgroup$ – Shubham Johri Dec 6 '19 at 8:12
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    $\begingroup$ @ShubhamJohri It is still a local maximum. All you need for a local maximum at is that $f(x)\le f(3)$ for an open interval containing 3. The inequality does not have to be strict (for $x\ne3$). It may seem odd, but a constant function has a local maximum at every point. $\endgroup$ – almagest Dec 6 '19 at 8:44

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