Question

If f(x)=k (square root)2+x, and f^-^1 (-15)=7, what is the value of k

Answer 1

There's a bit of ambiguity in your question...

We know that
`f^(-1)(-15)=7`, which means
`f(7)=-15`.

I see three possible interpretations:

• If
`f(x)=k\sqrt2+x`, then

`f(7)=-15=k\sqrt2+7\implies k\sqrt2=-22\implies k=-(22)/(\sqrt2)=11\sqrt2`

• If
`f(x)=k√(2+x)`, then

`f(7)=-15=k√(2+7)\implies -15=3k\implies k=-5`

• If
`f(x)=\sqrt[k]{2+x}`, then

`f(7)=-15=\sqrt[k]{2+7}\implies-15=9^(1/k)\implies\frac1k=\log_9(-15)`

which has no real-valued solution.

I suspect the second interpretation is what you meant to write.