$f (x) = | x - 1 | + | x + 2 | + | x - 3 |$ is not differentiable at

2 points

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 points

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 points

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 point

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 3 points
Given $f (x) = | x - 1 | + | x + 2 | + | x - 3 |$
$∴ f (x) = (1 - x) + - (x - 2) + (3 - x)$ For $x < - 2$
$= 2 - 3 x$ For $x < - 2$
$f (x) = (1 - x) + (x + 2) + (3 - x)$ For $- 2 \leq x < 1$
$= 6 - x$ For $- 2 \leq x < 1$
$f (x) = x - 1 + x + 2 + 3 - x$ For $1 \leq x < 3$
$= 4 + x$ For $1 \leq x < 3$
$f (x) = x - 1 + x + 2 + x - 3$
$= 3 x - 2$ For $x \geq 3$
$∴ f (x) = {2 - 3 x x < - 2$
${6 - x - 2 \leq x < 1$
${4 + x 1 \leq x < 3$
${3 x - 2 x \geq 3$
$f^{'} (x) = {- 3 x < - 2$
${- 1 - 2 \leq x < 1$
${1 1 \leq x < 3$
${3 x \geq 3$
$∴ f (x)$ is not differentiable at x=-2,1,3

Suggest Corrections

Similar questions

f (x) = (1 - x) ∣ ∣ x - x^{2} ∣ ∣ + x

f (x) = 2 + \sqrt{1 - x^{2}}, | x | \leq 1

f (x) = 2 e^{(1 - x)^{2}}

| x | > 1

f (x)

f (x) = 2 x + 1; x \leq 1

= x^{2} + 2; 1 < x \leq 2

= 4 x^{2} + 2; x > 2

f (x)

f (x) = | 2 x + 1 | - 3 | x + 2 | + | x^{2} + x - 2 |, x \in R

f (x) = | 2 x + 1 | - 3 | x + 2 | + | x^{2} + x - 2 |, x \in R

Join BYJU'S Learning Program