Vtyjkyuk

Pretty self explanatory. I'm trying to find the range of $$f(x)=arctan(1+frac1x)$$

but in all honesty, I'm not really sure how to proceed. I feel like there is something very silly and obvious I'm missing. I can find the domain fairly easily but how do I go about finding the range without taking its inverse? Is that even possible?

I have the same problem with:

$$f(x)=e^x+sqrtx^2+1$$

In general, how would I go about finding the range? I know I can just take the inverse and find the domain of that, but is that the only way? Thanks!

At first we need to show the range for $f(x)=1+frac1x$ which is of course $(-infty,infty)setminus1$, indeed

$$y=1+frac1x iff x=frac 1 y-1$$

therefore we can reach any value but not $y=1$.

For the second one we have that $x+sqrtx^2+1>0$ and

• $lim_xto infty x+sqrtx^2+1=infty$




• $lim_xto -infty x+sqrtx^2+1=lim_uto infty -u+sqrtu^2+1=lim_uto infty frac-u^2+u^2+1u+sqrtu^2+1=0$

In order to avoid limits note that

$$y=x+sqrtx^2+1 iff (y-x)^2=x^2+1 iff y^2-2xy+x^2=x^2+1 iff x=fracy^2-12y$$

Second is also the same as the first one you need to minimize $x+sqrtx^2+1$ in order to minimize $e^x+sqrtx^2+1$ bcause $e^x$ is monotonic increasing

$$lim_xto infty x+sqrtx^2+1=infty$$
$$lim_xto -infty x+sqrtx^2+1=0$$

$$fracddx(x+sqrtx^2+1)=1+fracxsqrtx^2+1=0$$
$$fracxsqrtx^2+1=-1$$
When $x>0$ this is not possible because LHS is positive and RHS is negative
Let's check when $x<0$

$$x^2=x^2+1$$ gives us $0=1$

Thus $(x+sqrtx^2+1)$ never becomes 0 it is always greater than 0

Thus $$e^(x+sqrtx^2+1)>1$$