(a) Clearly F is continuous on `(-oo,0),(0,oo)` since g and h are continuous. The only question is at 0.

Since h is continuous, `lim_(x->0^-)=h(0)`

Since g is continuous `lim_(x->0^+)=g(0)`

But h(0)=F(0)=g(0). So F is continuous at 0 since the limit at 0 exists and equals the value of F(0).

(b)...

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(a) Clearly F is continuous on `(-oo,0),(0,oo)` since g and h are continuous. The only question is at 0.

Since h is continuous, `lim_(x->0^-)=h(0)`

Since g is continuous `lim_(x->0^+)=g(0)`

But h(0)=F(0)=g(0). So F is continuous at 0 since the limit at 0 exists and equals the value of F(0).

(b) It is possible for g'(0) and h'(0) to exist, and for F to not be differentiable at 0. Consider h(x)=-x, g(x)=x. Then h'(0)=-1, g'(0)=1 and h(0)=0=g(0), and F is not differentiable since the derivative from the left is not equal to the derivative from the right.

In order to guarantee F differentiable, not only must h'(0),g'(0) exist, but h'(0) must equal g'(0).