Question

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

Answer 1

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.)

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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.