For simple ones, like quadratic equations, I can usually find the minimum point and give a correct a

Question

For simple ones, like quadratic equations, I can usually find the minimum point and give a correct answer.
But take, for example:

f (x) = \frac{3}{2 x + 1}, x > 0

and

g (x) = \frac{1}{x} + 2, x > 0

I am so confused with the whole process of finding the ranges of functions, including those above as samples, that I can't even quite explain what or why.
Can someone please, in a very step-by-step process, detail exactly what steps you would take to get the ranges of the above functions? I tried substituting x-values (such as 0), and came up with

f (x) > 3

, but that was mostly guesswork -- also,

f (x) > 3

is incorrect.
Also, is there an outline I can follow -- even for the thinking process, like check if A, check if B -- that would work every time?

Answer 1

I added an additional explanation for the limits part at the end.For f you have that x>0⇒2x>0⇒2x+1>1⇒12x+1<1⇒32x+1<3. On the other hand, 2x+1>1>0⇒12x+1>0⇒32x+1>0. So 00. Furthermore, limx→0+f(x)=3 and limx→+∞f(x)=0, so, since f is continuous in (0,+∞), we have that the range of f is all of (0,3).For g is similar. We have that for x>0⇒1x>0⇒1x+2>2. So g(x)>2 for every x>0. Furthermore, limx→0+g(x)=+∞ and limx→+∞g(x)=2, so, since g is continuous in (0,+∞), we have that the range of g is all of (2,+∞).I've been told that maybe I shouldn't use limits to justify the ranges, so I'll explain a little bit what I would say with words in the last part of the first paragraph if I didn't use limits.For f(x)=32x+1 we have that 00. Now, we can see that, as x goes closer and closer to 0 then the 2x part of the denominator starts to get smaller and smaller, almost vanishing. If we could take x=0 then we would get 3 as a value, but since we can't we can just say that the values of f keep getting closer and closer to 3; thus the parenthesis for the 3 in the range.If we make x go bigger and bigger then 2x+1 gets bigger and bigger, making the fraction 32x+1 go smaller and smaller, almost vanishing into 0. It never actually becomes 0, but it keeps getting closer from above (from the positive numbers); thus the parenthesis for 0 in the range. f is a good function in (0,+∞): there are no "explosions" in the denominator, and it's defined always by the same expression.This is not sufficient to say that f is continuous, but it does look good. If you have the tools or the information so you can assert f is indeed continuous, use it, since then f can't jump from one value to another one, it has to go continuously from one to the other, taking all values in between. This is the same as thinking that if you want to get to a certain height from the ground, you'll have to be at some moment at ever height in between. Then f takes every value between 0 and 3, since it can be as close to those values as we want.Now you can try this reasoning with g.

For simple ones, like quadratic equations, I can usually find the minimum point and give a correct a

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