Let f (x) = a + b |x| + c |x|4, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if

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(b) b = 0 We have, f\left( x ight) = a + b\left| x ight| + c \left| x ight|^4 `f(x) = {(a +bx +c|x|^4 ,xge 0),(a-bx +cx^4,x<0):}` 	ext { Here }, f\left( x ight)	ext { is differentiable at x = 0 } 	herefore \left(	ext { LHD at x } = 0 ight) = \left(	ext { RHD at x }= 0 ight) \Rightarrow \lim_{x 	o 0^-} \frac{f\left( x ight) - f\left( 0 ight)}{x - 0} = \lim_{x 	o 0^+} \frac{f\left( x ight) - f\left( 0 ight)}{x - 0} \Rightarrow \lim_{x 	o 0^-} \frac{a - bx + c x^4 - a}{x} = \lim_{x 	o 0^+} \frac{a + bx + c x^4 - a}{x} \Rightarrow \lim_{h 	o 0} \frac{a - b\left( 0 - h ight) + c \left( 0 - h ight)^4 - a}{0 - h} = \lim_{h 	o 0} \frac{a + b\left( 0 + h ight) + c \left( 0 + h ight)^4 - a}{0 + h} \Rightarrow \lim_{h 	o 0} \frac{a + bh + c h^4 - a}{- h} = \lim_{h 	o 0} \frac{a + bh + c h^4 - a}{h} \Rightarrow \lim_{h 	o 0} bh + c h^4/- h = \lim_{h 	o 0} bh + c h^4/h \Rightarrow \lim_{h 	o 0} \left( - b - c h^3 ight) = \lim_{h 	o 0} \left( b + c h^3 ight) \Rightarrow - b = b \Rightarrow 2b = 0 \Rightarrow b = 0

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