Is this true :if $x\in [0.2] $ then $f(x)=\frac{2x+3}{x+2} \in [0.2]$?

Question

I'm sorry to ask this question mayeb it's a trivial question but i would like to confirme if i have this function $f(x)=\frac{2x+3}{x+2}$ which $x$ is a real number in $[0.2] $ then $f(x) \in [0.2]$ ?

My question here is :

Is this true :if $x\in [0.2] $ then $f(x) \in [0.2]$ and why is not in $]0.2]$ ?

Note : The function is increasing and no values in the above area satisfy $f(x)=0$

Answer 1

$f(x)=\frac{2x+3}{x+2}=2-\frac{1}{x+2}$ is increasing on $[0,2]$, hence $f(x)\geq f(0)=\frac{3}{2}$ and $f(x)\leq f(2)=\frac{7}{4}$ for $x\in [0,2]$.

Answer 2

Hint:

$$\frac{2x+3}{x+2}=\frac{(x+2)+(x+1)}{x+2}=1+\frac{x+1}{x+2}$$

Answer 3

You only need to show that if $0\leq x\leq 2$ then $$0\leq f(x)\leq 2$$ but it's not needed that $f(x)=0$ for some $x\in [0,2]$.

Since $$f(x)=\frac{2x+3}{x+2}=\frac{2x+4-1}{x+2}=2-\frac{1}{x+2}\qquad\qquad\text{for all }x\neq -2$$ It follows that $$0\leq x\leq 2\quad\implies \quad2\leq x+2\leq 4\quad\implies\quad \frac14\leq \frac1{x+2}\leq\frac12$$ Then $$2-\frac12\leq2-\frac1{x+2}\leq2-\frac14\qquad\implies\qquad \frac32\leq f(x)\leq\frac74\quad \forall x\in[0,2]$$

Since $[\frac32,\frac74]$ is a subset of $[0,2]$ the result follows.