Fórum de Matemática
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If the Range of the function \(\displaystyle f(x) = \frac{x^2+ax+b}{x^2+2x+3}\) is \(\left[-5\;, 4\right]\)

then find value of \((a^2+b^2)\). Where \(a\;, b\in \mathbb{N}\)

We can see that \(x^2+2x+3\neq0 : \forall x \in \R\)

if the range of \(f(x)\) is between -5 and 4 and because we have a continuous function where the denominator is never zero, and

we see that \(\lim_{x \to +\infty}f(x)=\lim_{x \to -\infty}f(x)=1\)

we can conclude that \(x^2+ax+b\) will have always two roots and f(x) will have a maximum at \(f(x_1)=4\) and a minimum at \(f(x_2)=-5\)

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\(f(x_1)=4\) means that

\(\frac{x^2+ax+b}{x^2+2x+3}=4 \Leftrightarrow \frac{x^2+ax+b-4x^2-8x-12}{x^2+2x+3}=0\)

\(3(x_1)^2+(a-8)x_1+(b-12)=0\)

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\(f(x_2)=-5\) means that

\(\frac{x^2+ax+b}{x^2+2x+3}=-5 \Leftrightarrow \frac{x^2+ax+b+5x^2+10x+15}{x^2+2x+3}=0\)

\(6(x_2)^2+(a+10)x_2+(b+15)=0\)

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Realize also that

\(f'(x_1)=f'(x_2)=0\)

\(f'(x)=\frac{(2x+a)(x^2+2x+3)-(2x+2)(x^2+ax+b)}{(x^2+2x+3)^2}=\frac{(2-a)x^2+(6-2b)x+3a-2b}{(x^2+2x+3)^2}\)

Because the denominator is always positive solving \(f'(x)=0\) means that we just consider solving

\((2-a)x^2+(6-2b)x+3a-2b=0\)

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Remeber also because \(x^2+ax+b\) has two roots that \(a^2>4b\) and because \(a,b \in \mathbb{N}\) we see that \(a>2\sqrt{b}\)

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So, we now know that:

\(\begin{cases} 3(x_1)^2+(a-8)x_1+(b-12)=0 \\ (2-a)(x_1)^2+(6-2b)x_1+3a-2b=0 \end{cases}\)

\(\begin{cases} 6(x_2)^2+(a+10)x_2+(b+15)=0 \\ (2-a)(x_2)^2+(6-2b)x_2+3a-2b=0 \end{cases}\)

I am not clearly realizng how to solve everything but from this step you may try to achive what you pretend

Best regards and please share any results

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João Pimentel Ferreira
 
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