Answer
$$k(x)=1-\frac{1}{x-1}$$
The graph is shown below.
![](https://gradesaver.s3.amazonaws.com/uploads/solution/3a4ddb8e-cf8a-43b6-9d42-bece0b12ffcb/result_image/1536660798.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T014447Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=2d2b1f232d389f3ab3592cc797f50e260252d2ae73b04bc3f1f3a32839480bbf)
Work Step by Step
(What I present here is just an example. There are other possible functions which satisfy the exercise.)
- As $\lim_{x\to1^-}k(x)=\infty$ and $\lim_{x\to1^+}k(x)=-\infty$, we would suppose the function $k(x)$ to be in the form of a rational function $k(x)=A/B$ where $B=x-1$ but $A\lt0$ so that we would have the reverse $\lim_{x\to1^-}k(x)=\infty$ and $\lim_{x\to1^+}k(x)=-\infty$.
$$k(x)=\frac{A}{x-1}$$
- As $\lim_{x\to\pm\infty}k(x)=1$
$$\lim_{x\to\pm\infty}\frac{A}{x-1}=1$$
We can set $\lim_{x\to\pm\infty}\frac{A}{x-1}=0$ then add $1$ following. In detail, $$\lim_{x\to\pm\infty}\frac{A}{x-1}+1=1$$
Any negative numbers would suit $A$ now. I would pick $A=-1$
In conclusion, the function I come up with is $$k(x)=1-\frac{1}{x-1}$$
The graph is shown below.
![](https://gradesaver.s3.amazonaws.com/uploads/solution/3a4ddb8e-cf8a-43b6-9d42-bece0b12ffcb/steps_image/small_1536660798.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T014447Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=aa69df31b1ece448afcf2fac36329979a92fa2a817ff2832cc4ca9893bc30daf)