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

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1 Answer
Truong-Son N.
Mar 14, 2017

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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