If \frac{1}{g\left(x\right)}=k has exactly 3 solutions, k cannot be zero since \frac{1}{\text{anything}} is never 0. Thus we can take the reciprocal of both sides, and say that

g\left(x\right)=\frac{1}{k}

has 3 solutions. This means that if we draw a horizontal line at y=\frac{1}{k}, it intersects y=g\left(x\right) exactly 3 times.

From the sketch above, we see that \frac{1}{k} must be between the horizontal asymptote and the local maximum of g:

The horizontal asymptote of g can be found using the fact that f\left(x\right) had an asymptote at y=-20, which means that g has an asymptote at

y=2\cdot\left|-20\right|-38=2

Thus we have

2<\frac{1}{k}<34

Notice that since \frac{1}{k} is positive, k is positive also.

Consider each piece:

2<\frac{1}{k}\Rightarrow2k<1\Longrightarrow k<\frac12

and

\frac{1}{k}<34\Rightarrow34k>1\Rightarrow k>\frac{1}{34}

Putting the two together:

\boxed{\frac{1}{34}<k<\frac12}

Problem Bank - Exponents & Logarithms

Problem Bank - Exponents & Logarithms