Let the functions f: R rightarrow R and g: R | vedclass

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(B) First,we find the composite function $(f \circ g)(x) = f(g(x))$.Given $g(x) < 0$ when $x < 0$ (since $x^3 < 0$ for $x < 0$),and $g(x) \geq 0$ when

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(B) First,we find the composite function $(f \circ g)(x) = f(g(x))$.Given $g(x) Thus,$(f \circ g)(x) = \begin{cases} g(x)+2, & g(x) Now,check for continuity and differentiability at the transition points $x=0$ and $x=1$.At $x=0$: $\lim_{x 	o 0^-} (x^3+2) = 2$ and $\lim_{x 	o 0^+} (x^6) = 0$. Since $2 
eq 0$,the function is discontinuous at $x=0$,hence not differentiable at $x=0$.At $x=1$: $\lim_{x 	o 1^-} (x^6) = 1$ and $\lim_{x 	o 1^+} (3x-2)^2 = (3(1)-2)^2 = 1$. The function is continuous at $x=1$.Check derivatives at $x=1$: $LHD = \frac{d}{dx}(x^6)|_{x=1} = 6(1)^5 = 6$. $RHD = \frac{d}{dx}(3x-2)^2|_{x=1} = 2(3x-2) \cdot 3|_{x=1} = 6(3-2) = 6$.Since $LHD = RHD$,the function is differentiable at $x=1$.Therefore,the only point where $(f \circ g)(x)$ is not differentiable is $x=0$. The number of such points is $1$.

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