If $$a,b$$ are real numbers such that the fraction function $$y = \dfrac{ax+b}{x^2+1}$$ has minimum value -1 and maximum value 4, find the value of $$a,b$$

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If $$a,b$$ are real numbers such that the fraction  function $$y = \dfrac{ax+b}{x^2+1}$$  has minimum value -1 and maximum value 4, find the value of $$a,b$$

Uniteasy Admin · Accepted Answer

$$y = \dfrac{ax+b}{x^2+1}$$
$$ax+b=yx^2+y$$
A quadratic equation was obtained in terms of $$x$$
$$yx^2-ax+y-b=0$$
For a quadratic equation to have real solutions, 
the discriminant must be greater than or equal to $$0$$, that is,
$$\Delta =a^2-4y(y-b)\geq  0$$
$$ -4y^2+4by+a^2\geq 0$$
$$y^2-by-\dfrac{a^2}{4} \leq 0$$n
$$y = \dfrac{ax+b}{x^2+1}$$ has minimum value -1 and maximum value 4, that is
$$(y-4)(y+1) \leq 0$$
Expand the inequality
$$y^2-3y-4\leq 0$$n
Comparing (2) with (1) results in the system of equations in terms $$a,b $$
$$\begin{cases} a &=3 \ -\dfrac{a^2}{4}&=-4 \end{cases}$$
Solving for $$a,b$$ yields 
$$ b=3, a=\pm4 $$

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