Proving m<f(x)<M for f(x)=\frac{x}{x^2 +x+1}

[Andrax]

Homework Statement

prove that m<f(x)<M
so $\forallx$$\in$R: f(x)=$\frac{x}{x^2 +x+1}$
the question is prove that m$\leq$f(x)$\leq$M
M and m are real numbers

Homework Equations

The Attempt at a Solution

all i did so far was making f(x) : (x+1)^2 +(3/4) well i noticed that m=-1 and M=4/3 from the graph but i can't really prove it well
f(x)= 1-$\frac{x^2+1}{x^2 +x +1}$ = 1-$\frac{x^2+1}{(x+1/2)^2+3/4}$

 

[Michael Redei]

If you know the values of m = -1 and M = 4/3, you can try a direct approach:

If you want to prove $-1 \leq \frac x{x^2+x+1}$, you can multiply both sides of that inequality by $(x^2+x+1) = (x+\frac12)^2+\frac34$, which is positive (and so will not change the direction of the $\leq$ relation symbol):
$$
-(x^2+x+1) \leq \frac{x(x^2+x+1)}{x^2+x+1}.
$$
Can you now prove that this inequality is true?

 

[Andrax]

If you know the values of m = -1 and M = 4/3, you can try a direct approach:

If you want to prove $-1 \leq \frac x{x^2+x+1}$, you can multiply both sides of that inequality by $(x^2+x+1) = (x+\frac12)^2+\frac34$, which is positive (and so will not change the direction of the $\leq$ relation symbol):
$$
-(x^2+x+1) \leq \frac{x(x^2+x+1)}{x^2+x+1}.
$$
Can you now prove that this inequality is true?

well there is a slight problem here the exercise dosen't mention the -1 nor the 4/3
so they are expecting us to do it "manually".

 

[pasmith]

One approach to this sort of problem is to ask "What conditions must $y$ satisfy for $y = f(x)$ to have real solutions for $x$?"

This is a particularly good approach in this case, because if $y = x/(x^2 + x + 1)$ then
$$yx^2 + (y-1)x + y = 0$$
and hopefully you know the condition for that quadratic in $x$ to have real roots. That will give you a condition which $y$ must satisfy, which in turn will give you bounds for $f(x)$.