Find this Question value using range method.

Question

If the range of this function $f(x)$=$\frac{x-A}{x^2-5x+6}$ = R. Then, solve for A .

How can I solve this question.

Answer 1

Let

$$f_A(x):=\frac{x-A}{x^2-5x+6}=\frac{x-A}{(x-2)(x-3)}=R\tag{1}$$

The question is: For a fixed value of $A$, being given value of $R$, can it be reached as image by $f_A$ of a certain real value of $x$ ?

Otherwise said, for a fixed value of $A$, what is $f_A(\mathbb{R} \backslash \{2,3\})$ ?

A particular case that can be treated apart: the case where $M=0$ which is evidently reached for any value of $A$ but values $A=2$ and $A=3$ (due to simplification by $(x-2)$ or $(x-3)$).

If we plot the curves of $f_A$ for different values of $A$, one sees that there are different cases:

Fig. 1: Curves of $y=f_A(x)$. Cases $A<2$ in red; cases $2<A<3$ in black ; cases $A=2$ and $A=3$ in magenta and cyan resp.; cases $A \ge 3$ in blue.

There is an evident symmetry with respect to point $(2.5;0)$. In order to take profit of this important remark, let us set:

$$g_B(X)=f_A(X+\frac{5}{2})=\dfrac{X-B}{X^2-\tfrac14} \ \text{with} \ B:=A-\tfrac52$$

We have now to find, for a fixed $B$, the range of values of $M$ for which there exists $X$ such that:

$$g_B(X)=\dfrac{X-B}{X^2-\tfrac14} =M \ \ \iff \ \ MX^2-X+(B-\tfrac14M)=0$$

Otherwise said, we have to "discuss", according to the values of $M$ if this quadratic equation has solutions which is equivalent to the fact that its discriminant is $\ge 0$; i.e.,

$$\underbrace{M^2-4MB+1}_{q_B(M)} \ge 0\tag{2}$$

Due to the symmetry of the issue, it is sufficient to consider positive values of $M$.

Let $\delta=4B^2-1$ be the discriminant of $q_B(M)$.

It is known that the sign of such a quadratic expression is

• always $>0$ if $\delta<0$ i.e., if $|B|<1/2$. This means that function $g_B$ is surjective in this case (and only in this case).

• $>0$ if

$$M \ \text{is outside interval} \ [\underbrace{2B-\sqrt{4B^2-1}}_{M_1};\underbrace{2B+\sqrt{4B^2-1}}_{M_2}]$$

where $M_1,M_2$ are the roots of quadratic $q_B$.

It remains to express these conditions in terms of $A$.

Remark: Verification in the case of $B=\frac34 \ \iff A=\frac{13}{4}=3.25$: in this case $[M_1;M_2]=[\tfrac{3-\sqrt{5}}{2};\tfrac{3+\sqrt{5}}{2}]\approx[0.382;2.618]$ corresponding to the range of values not "covered" by the blue curve above all others in the region $x>3$ of the figure.