Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.

Question

Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.

My attempt:

We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.

then for $a\geq 1$:

so then $$2-a=0\implies a=2$$ and $$\frac {a+2}3=2\implies a=4.$$

and for $a\leq1:$

so then $$2-a=3\implies a=-1$$ and $$\frac {a+2}3=0\implies a=-2.$$

Now my answers are in the type of interval. How do I know which interval to choose?

Answer 1

The derivative of $f(x)$ is equal to $\dfrac{(a-1)(x^2-1)}{(x^2+x+1)^2}$ so the extremes are independent of a and taken at $x=\pm 1$. This extremes are $\dfrac{2\pm a}{3}$ It follows $0\le a\le 2$

Answer 2

Lemma: If $x^2+mx+n\geq 0$ for all $x$ then discriminant $m^2-4n \leq 0$.

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First note that $x^2+x+1>0$ for all $x$ (since discriminant =$-3$)

$$0\leq f(x)\leq 2 \implies 0\leq x^2+ax+1\leq 2x^2+2x+2$$

• From $0\leq x^2+ax+1$ we get $a^2-4\leq 0$ so $|a|\leq 2$ so $\boxed{-2\leq a\leq 2}$.
• From $x^2+ax+1\leq 2x^2+2x+2$ we have $0\leq x^2+(2-a)x+1$ so $(a-2)^2-4\leq 0$ so $|a-2|\leq 2$ so $\boxed{0\leq a\leq 4}$.

Thus $a\in[0,2]$.

Answer 3

Rewrite

$$f(x) = 1 + \frac{(a-1)x}{x^2+x+1}$$

and note that $f(x) \in [0,2]$ is equivalent to $(a-1)\frac{x}{x^2+x+1} \in [-1,1]$.

Notice that $\frac{x}{x^2+x+1} \in [-1,1]$:

$$\frac{x}{x^2+x+1} \le 1 \iff 0 \le x^2+1$$

$$\frac{x}{x^2+x+1} \ge -1 \iff 0 \le x^2+2x + 1$$

Now, if $a \in [0,2]$ then $|a-1| \le 1$ so

$$|a-1|\underbrace{\left|\frac{x}{x^2+x+1}\right|}_{\le 1} \le 1, \forall x \in \mathbb{R}$$

Conversely, if $(a-1)\frac{x}{x^2+x+1} \in [-1,1], \forall x \in \mathbb{R}$ then in particular for $x = -1$ we get $1-a \in [-1,1]$ which implies $a \in [0,2]$.

We conclude $f(x) \in [0,2], \forall x \in \mathbb{R}$ if and only if $a \in [0,2]$.

Answer 4

You're doing well. The derivative is $$ f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}=(1-a)\frac{(x^2-1)}{(x^2+x+1)^2} $$ Leaving aside $1-a$, the fraction is negative for $|x|<1$.

If $1-a>0$, the function has a maximum at $-1$ and a minimum at $1$. The situation is reversed for $1-a<0$. For $a=1$ the function is constant and satisfies the requirement.

Since $\lim_{x\to\pm\infty}f(x)=1$, we just need to compute $$ f(-1)=2-a \qquad f(1)=\frac{2+a}{3} $$ Thus we must have $$ \begin{cases} 1-a>0 \\[4px] 2-a\le2 \\[4px] 2+a\ge0 \end{cases} \qquad\text{or}\qquad \begin{cases} 1-a<0 \\[4px] 2-a\ge0 \\[4px] 2+a\le6 \end{cases} \qquad\text{or}\qquad a=1 $$ Solving this gives $0\le a\le 2$.