I'm not sure how to approach these types of conceptual questions - would I have to use proofs? Suppose g(x) and h(x) are two continuous functions (on all real numbers) such that g(0) = h(0). Then if we define a function F by: F(x) = g(x) if x >= 0, h(x) if x <=0 (a) show that the function F is continuous for all real numbers. (b) If g′(0) and h′(0) exists, is it enough to say that F(x) is differentiable at x = 0? If not, give an example of such a case, i.e., give a function g(x) and another function h(x) such that g(0) = h(0), and g′(0) and h′(0) exist, but F′(0) does not. What condition will you then need for g′(0) and h′(0), such that F′(0) exists?  

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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)...

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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).

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