If f(x)=|x|+| sin x| for x ∈(-(π/2), (π/2)), then its left hand derivative at x=0 is

Question

If f(x)=|x|+| sin x| for x ∈(-(π/2), (π/2)), then its left hand derivative at x=0 is (A) 0 (B) -1 (C) -2 (D) -3

Tardigrade Admin · Accepted Answer

f(x)=|x|+| sin x| LHD = displaystyle lim h arrow 0 (f(0-h)-f(0)/0-h) = displaystyle lim h arrow 0 (|0-h|+| sin (0-h)|-(0+0)/0-h) = displaystyle lim h arrow 0 (h+ sin h/-h)=- displaystyle lim h arrow 0(1+( sin h/h)) =-(1+1)=-2

Q. If $f (x) = ∣ x ∣ + ∣ sin x ∣$ for $x \in (- \frac{π}{2}, \frac{π}{2})$ , then its left hand derivative at $x = 0$ is

0

-1

-2

-3

$f (x) = ∣ x ∣ + ∣ sin x ∣$
$L HD = h \to 0 lim \frac{f ( 0 - h ) - f ( 0 )}{0 - h}$
$= h \to 0 lim \frac{∣0 - h ∣ + ∣ sin ( 0 - h ) ∣ - ( 0 + 0 )}{0 - h}$
$= h \to 0 lim \frac{h + sin h}{- h} = - h \to 0 lim (1 + \frac{sin h}{h})$
$= - (1 + 1) = - 2$

Q. If f(x)=∣x∣+∣sinx∣ for x∈(−2π​,2π​), then its left hand derivative at x=0 is

0

-1

-2

-3

f(x)=∣x∣+∣sinx∣ LHD=h→0lim​0−hf(0−h)−f(0)​ =h→0lim​0−h∣0−h∣+∣sin(0−h)∣−(0+0)​ =h→0lim​−hh+sinh​=−h→0lim​(1+hsinh​) =−(1+1)=−2

Q. If $f (x) = ∣ x ∣ + ∣ sin x ∣$ for $x \in (- \frac{π}{2}, \frac{π}{2})$ , then its left hand derivative at $x = 0$ is

$f (x) = ∣ x ∣ + ∣ sin x ∣$
$L HD = h \to 0 lim \frac{f ( 0 - h ) - f ( 0 )}{0 - h}$
$= h \to 0 lim \frac{∣0 - h ∣ + ∣ sin ( 0 - h ) ∣ - ( 0 + 0 )}{0 - h}$
$= h \to 0 lim \frac{h + sin h}{- h} = - h \to 0 lim (1 + \frac{sin h}{h})$
$= - (1 + 1) = - 2$