Finding the range of a function, which contains $a$ and its multiplicative inverse

Question

Finding the range of a function seems to be generally an easy task, but this time I doubt the way I solved the problem to find the range.

The problem:

Find the range the range of the function below $$f(x)=sin^2(x) + csc^2(x)$$

My attempt:

We know that the range of $sin^2(x)$ is $[0;1]$ and that the one of $csc^2(x)$ is $[1;\infty[$ Knowing this, we can model a simpler function which should have the same range, such a function will be: $$f(x)=x^2+1/x^2, x \in [0;1]$$

Taking the derivative to find the minimum or maximum $$f'(x)=2x-2/x^3, x \in [0;1]$$ $$0=2x-2/x^3, x \in [0;1]$$ $$x=1$$ $$f(1)=2$$ Which is a local minima, and the only one given the restriction. Thus the range of both function must be $[2; \infty[$

Answer 1

Your answer is correct. There is a hole in the logic when you transform $\sin^2 x + \csc^2 x$ to $x^2 + 1/x^2$ because you only cite the ranges of each function. It is critical that $\sin^2 x = 1/\csc^2 x$ for this to work but you don't say that. If we were considering $f(x)+g(x)$ where the range of $f(x)$ is $[0,1]$ and the range of $g(x)$ is $[1,\infty)$ it could be that $f(x)=0$ when $g(x)=1$ and the sum could be as low as $1$. It would also be better not to reuse $x$ here. There are lots more letters.

Answer 2

$$f(y) = y + \frac1{y}$$

where $ 0 < y \le 1 $

$$f'(y)=1-\frac1{y^2}$$

Hence it is decreasing when $y<1$.

Also $\lim_{y \to 0^+}f(y) =\infty$

Hence the range is $[2,\infty)$.

Answer 3

One more answer:

$f(x)=x^2 +1/(x^2)$, $x^2 \in (0,1]$.

AM-GM:

$f(x)= x^2+1/(x^2) \ge 2\sqrt{(x^2) \cdot 1/(x^2)} =2$.

Equality for $x^2=1/(x^2) =1$.

Considering $x^2 \rightarrow 0$, and since f is continuos,

we get : $R_f = [2,\infty)$.