Finding suitable variables in a real number inequality

Question

I have some trouble with a inequality problem.

I have to find two values that will work in the given inequality. The problem goes as follows:

Let $0 < x <2.$ Find real numbers $1 < a < b < 2,$ so that

$$a \leq \frac{(x+3)}{(x+2)} \leq b.$$

I tried to solve this problem by multiplying the whole function with $(x+2)$, but that didn't work since then $a = 1$, which is wrong.

Any and all help is very appreciated!

Answer 1

Let $f(x) = \frac{x+3}{x+2}$, take the derivative we get $f'(x) = -\frac{1}{(x+2)^2}$. Notice $f'(x) < 0$ for all $x$ in it's domain, which tells us that $f(x)$ is a decreasing function.

We can now say that the maximum of $f(x)$ on the domain $0 < x < 2$ will occur when $x=0 \rightarrow f(0)=\frac{3}{2} = 1.5$ similarly minimum occurs when $x=2 \rightarrow f(2) = \frac{5}{4} = 1.25$

We can therefore say, no matter the value of $x$, $1.25 \leq \frac{x+3}{x+2} \leq 1.5$.

Here is a Desmos graph to help you visualize.

Hope this helps!

Answer 2

Hints

We have $$\frac{x+3}{x+2}=1+\frac{1}{x+2}$$ and the function is decreasing on $(0,2)$.