# A) f(x)= 6x+7x^(11) find f^(-1)(-13)= ? find (f^(-1))'(-13)= ? B) g(x)=x^(2)-8x+22 for x is an element of [4, positive infinity) find g^-1(7)= ? find (g^-1)'(7)=?

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### 1 Answer

(1) We are given that `f(x)=6x+7x^11` .

(a) We are asked to find `f^(-1)(-13)` . One way would be to compute the inverse function and evaluate at x=-13. This is very difficult. It is much easier to realize that for an inverse function; `f(f^(-1)(x))=f^(-1)(f(x))=x` .

Here, we can find an x such that f(x)=-13. We get `6x+7x^11=-13` . By inspection we have x=-1, so the point (-1,-13) lies on the graph of f(x). Thus the point (-13,-1) lies on the graph of `f^(-1)(x)` .

**The value of** `f^(-1)(-13)=-1`

(b) In order to find `(f^(-1))'(-13)` we use the fact that `(f^(-1))'(x)=1/(f'(f^(-1)(x)))` . With `f'(x)=6+77x^10` we get:

`(f^(-1))'(-13)=1/(6+77(-1)^10)=1/83`

**So the value of** `(f^(-1))'(-13)=1/83`

(2) We are given `g(x)=x^2-8x+22` , for x>4.

(a) Again it is difficult to find `g^(-1)(x)` . We find ` ` :

`x^2-8x+22=7`

`x^2-8x+15=0`

`(x-5)(x-3)=0`

`=>x=5` or `x=3` . x=3 is not in the domain, so we have the point (5,7) on g(x). Thus the point (7,5) lies on the graph of `g^(-1)(x)` .

**Thus the required value is** `g^(-1)(7)=5`

(b) Using the theorem cited above, and `g'(x)=2x-8` we get:

`(g^(-1))'(7)=1/(2(5)-8)=1/2`

**Then the required value is** `(g^(-1))'(7)=1/2`

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