If f x -1 for x- 0-1-sin x for 0- x- - -2- then at x-0- then show that the derivative f- x does not exist-

#160;We have f ( 0 )=1+sin 0=1. #160;RHL #160; Rf' ( 0 )=lim_h→ 01+sin ( 0+h ) 1/h=lim_h→ 0sin h/h=1 #160;and #160;LHL Lf' ( 0 ) ...

If f x =1 f...

Question

If $f (x) = 1$ for $x < 0 = 1 + sin x$ for $0 \leq x < π / 2,$ then at x=0, then show that the derivative $f^{'} (x)$ does not exist.

Open in App

Solution

Suggest Corrections

Similar questions

f (x) = {\begin{matrix} 1, & x < 0 1 + sin x, & 0 \leq x < \frac{π}{2} \end{matrix}

x = 0

f^{'} (x)

f (x) = \frac{t a n π [(x - π)]}{1 + [x]^{2}}

f (x) = | x | + | sin x |

x \in (- \frac{π}{2}, \frac{π}{2})

x = 0

lim x \to 0 \frac{\sqrt{1 - cos 2 x}}{x}

| sin x | = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} sin x; & 0 < x < \frac{π}{2} - sin x; & - \frac{π}{2} < x < 0 \end{matrix}

f (x) = \frac{t a n (π [x - π])}{1 + [x]^{2}}