Question

Question #b61f5

Answer 1

`G'(2)=-1`.

Explanation:

Knowing that, `g` is the inverse function of `f`, we have,

`g@f"=the Identity fun. "rArr g(f(x))=I(x)=x....(star)`

`rArr d/dx g(f(x))=d/dx(x).`

`"By the Chain Rule, "g'(f(x))f'(x)=1.`

In particular, for `x=3, g'(f(3))f'(3)=1................(ast)`

But, `f(3)=2, and, f'(3)=1/9 :. (ast) rArr g'(2)=9..................(1)`

Now, `G(x)=1/g(x)rArr G'(x)=-(g'(x))/[g(x)]^2..."[the Chain Rule]"`

Hence, for `x=2, G'(2)=-(g'(2))/[g(2)]^2=-9/[g(2)]^2, ........[because, (1)]`

`"Here, to find, "g(2)," we take "x=3" in "(star)" & get, "g(f(3))=3, i.e., g(2)=3," as "f(3)=2.`

So, finally, `G'(2)=-9/3^2=-1.`

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