What are the left-hand and right-hand derivatives of $f(x) = [x] \sin \pi x$ at $x = 1$ ?

0

and

-\pi

Correct answerYour answer

-\pi

and

-\pi

Correct answerYour answer

\pi

and

-\pi

Correct answerYour answer

0

and

\pi

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

Just left of 1, $[x] = 0$ , so $f(x) = 0$ there: the left-hand derivative is 0.
At 1 and just right of it, $[x] = 1$ , so $f(x) = \sin \pi x$ , with slope $\pi \cos \pi = -\pi$ at 1.

Worth noting

$-\pi$ and $-\pi$ treats $[x]$ as 1 on both sides; to the left of 1 it is 0, which flattens $f$ .

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

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

What are the left-hand and right-hand derivatives of f(x)=[x]sin⁡πxf(x) = [x] \sin \pi xf(x)=[x]sinπx at x=1x = 1x=1?