Consider the function $f(x)=\frac{1}{x-1}+3^x-x^3-x^2.$ By inspection, $x=0$ is a root of $f(x)$. Show that $f(x)$ has at least three other roots and a vertical asymptote.
$\textbf{Solution:}$ This is not a hard question and I did fully solve it. Since $\displaystyle{\lim_{x \to 1^+}}f(x)=\infty$ and $\displaystyle{\lim_{x \to 1^-}}f(x)=-\infty$, we definitely know that $x=1$ is a vertical asymptote.
I'm not sure about how to show there are $3$ roots for $f(x)$. I know I can use the intermediate value theorem and show there is a sign change.
For example, I know that $f(0.05)=0.001$, $f(0.5)=-0.64$ so there is a root between $(0.05,0.5)$.
Similarly, $f(1.5)=1.57$, $f(2)=-2$ so there is a root between $(1.5,2)$.
$f(3)=-\frac{17}{2}$, $f(4)=\frac{4}{3}$ so there is a root between $(3,4)$.
So I've just proven that there at at least 3 roots.
The problem is that I basically used Wolfram alpha to get those intervals and use the Intermediate value theorem that way though. Is there a more nicer way of doing this that does not require me to use a calculator to get the intervals near that region? There should be some logical way I guess?
EDIT: Corrected $f(x)=\frac{1}{x-1}+3^x-x^2-x^2$ into $f(x)=\frac{1}{x-1}+3^x-x^3-x^2.$
Since $x=1$ cannot be a root, it may be simpler to consider $$g(x)=3^x (x-1)-(x^4-x^2+1)$$
By inspection $$g(-2)=-\frac{34}{3} \qquad g(-1)=\frac{1}{3}\qquad g(0)=0\qquad g(1)=1$$ $$ g(2)=-2 \qquad g(3)=-17 \qquad g(4)=4$$