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

$f(7)=-15=k\sqrt2+7\implies k\sqrt2=-22\implies k=-\dfrac{22}{\sqrt2}=11\sqrt2$

• If $f(x)=k\sqrt{2+x}$, then

$f(7)=-15=k\sqrt{2+7}\implies -15=3k\implies k=-5$

• If $f(x)=\sqrt[k]{2+x}$, then

$f(7)=-15=\sqrt[k]{2+7}\implies-15=9^{1/k}\implies\dfrac1k=\log_9(-15)$

which has no real-valued solution.

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