# Find #h'(2)#? (see image below)

The graphs above were:

Some things to keep in mind:

- Note that
#h(x) = f(g(x))# . This means you will have to find a value for#g(x)# , and use that value for the*argument*of#f(x)# . For instance, for some#g(3) = w# , we have that#h(3) = f(g(3)) = f(w)# . - We should recognize that the derivative at a
*corner*, e.g. when the graph abruptly changes slope, is*undefined*. #h'(x) = f'(g(x))*g'(x)# by the chain rule.Thus,

#h'(x)# does not exist if#g(x)# or#g'(x)# do not exist. (If#f'(x)# does not exist at some specified point, it doesn't necessarily imply that#g(x)# does not exist.)

Reading from the above graph:

For

#h'(2)# , we have that:#h'(2) = f'(g(2))g'(2)# #= f'(4)g'(2)# #= 3*-1# #= -3# Therefore,

#h'(2) = -3# , and it exists. Two of the question options apparently has us check#h'(-2)# , so let's see.#h'(-2) = f'(g(-2))g'(-2)# #= f'(1)g'(-2)# #= -1*-4# #= 4# Well, we found that

#h'(2)# exists... but neither of the answer choices in which it exists has the correct#h'(-2)# . Either there is a typo in the question or there is no correct answer choice.

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To find h'(2), we need to differentiate the function h(x) with respect to x and then evaluate it at x = 2. However, since there is no image provided, I am unable to determine the function h(x) and provide the accurate answer. Please provide the necessary information or equation for h(x) so that I can assist you further.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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