# Finding the range of a function analytically

I have a really simple question in need of a simple answer. Basically, when trying to find the range of any function, i just draw a graph and solve from there. I guess that's the usual thing to do. But, here's my question: Is there any other way to find the range, other than drawing the graph, let's say, analytically? If this sounds obvious to you, please respect the ingenuity of others not that gifted (like me). Thanks and have a great day!

• The graph is just for intuition, to rigorously prove it you shouldn't use graphs. There are exceptions depending on how you're assessed, but mathematically speaking, you need a little more. So yes, there are other ways. Jan 9, 2019 at 21:22
• In general finding the range of a function is not an easy task. In most elementary courses you are taught how to figure out the range using a graph. But what you are doing in that case is leaving the computational tasks to a computer. In a slightly more advanced class you will see that most functions a constructed from simpler functions and you can calculate the range using the information about those simpler functions. Jan 9, 2019 at 21:26

There is no "universal analytic algorithm". But here is an analytic example: Let $$f(x)=x^4-x^3+1,$$ with domain $$\Bbb R.$$
For $$x\ne 0$$ we have $$\frac {f(x)}{x^4}=1-\frac {1}{x}+\frac {1}{x^3},$$ so $$\lim_{|x|\to \infty}\frac {f(x)}{x^4}=1.$$ So $$f(x) >\frac {x^4}{2}$$ for all sufficiently large $$|x|$$.
Now for any $$z\in [f(0),\infty)$$ there exists $$r>0$$ such that $$f(r)>r^4/2>z.$$ Since $$f$$ is continuous we have $$\{f(t): t\in [0,r] \} \supset [f(0),f(r)] \supset \{z\}$$ so there exists $$t\in [0,r]$$ with $$f(t)=z,$$ so $$z\in ran (f).$$ Therefore $$(1)...\quad [f(0),\infty)\subset ran (f).$$ Now take $$s>0$$ such that $$f(t)>f(0)$$ whenever $$|t|\ge s.$$ Since $$f$$ is continuous and $$[-s,s]$$ is a closed bounded interval, there exists $$x_0 \in [-s,s]$$ such that $$f(x_0)=\min \{f(t): |t|\le s\}.$$ And we have $$f(x_0)\le f(0 ) whenever $$|t|>s$$. Therefore $$f(x_0)\le f(t) \text { for all } t\in \Bbb R.$$ And since $$f$$ is differentiable we must have $$0=f'(x_0)=4x_0^3-3x_0^2,$$ implying $$(x=0$$ or $$x_0=3/4).$$ But since $$f(0)>f(3/4)\ge f(x_0),$$ we must have $$x_0=3/4.$$
Since $$f$$ is continuous we have $$[f(3/4),f(0)]\subset \{f(t): t\in [0,3/4]\}.$$ Therefore $$(2)...\quad [f(3/4),f(0)]\subset ran (f).$$ Combining (1) and (2) we have $$(3)...\quad [f(3/4),\infty)\subset ran (f).$$ Finally, since no $$f(t)$$ can be less than $$f(3/4)=f(x_0),$$ we have $$(4)...\quad ran(f)\subset [3/4,\infty)$$ and from (3) and (4) we have $$ran (f)=[f(3/4),\infty).$$
Note: In the 3rd paragraph ("Now for any $$z\in [f(0),\infty)$$...") the choice to use $$f(0)$$ was arbitrary and not related to the fact that $$f'(0)=0.$$