Question

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

Collected in the board: Fractions

Steven Zheng posted 7 hours ago

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
(1)

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
(2)

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

Steven Zheng posted 7 hours ago

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