Question

If the range of the function $f ( x ) = \frac{x - 1}{p - x^2 + 1}$ does not contain any values belonging to the interval $[ - 1 , - \frac{1}{3} ]$, then what is the true set of values of $p$?

Answer 1

$(-\infty, -2] \cup [1/4, \infty)$

Answer 2

The function is given by $f(x) = \frac{x-1}{p-x^2+1}$. Let $y = f(x)$. $y(p-x^2+1) = x-1$ $yp - yx^2 + y = x - 1$ $yx^2 + x - (yp + y + 1) = 0$

This is a quadratic equation in $x$ for $y \ne 0$. For $x$ to be real, the discriminant must be non-negative. $\Delta = 1^2 - 4(y)(-(yp+y+1)) = 1 + 4y(yp+y+1) = 1 + 4y^2p + 4y^2 + 4y = 4py^2 + 4y^2 + 4y + 1$. For real $x$, we must have $4y^2 + (4p+4)y + 1 \ge 0$. Let $g(y) = 4y^2 + (4p+4)y + 1$. The range of $f(x)$ (for $y \ne 0$) is the set of $y$ such that $g(y) \ge 0$.

If $y=0$, the equation becomes $x - (0 \cdot p + 0 + 1) = 0$, which gives $x=1$. For $x=1$ to be in the domain of $f(x)$, the denominator $p-1^2+1 = p$ must be non-zero. If $p \ne 0$, $f(1) = \frac{1-1}{p-1+1} = 0/p = 0$, so $y=0$ is in the range. If $p=0$, $x=1$ is not in the domain.

The problem states that the range of $f(x)$ does not contain any values in the interval $[-1, -1/3]$. This means that for any $y \in [-1, -1/3]$, $y$ is not in the range of $f(x)$. The set of $y$ values for which $g(y) \ge 0$ must not intersect the interval $[-1, -1/3]$.

The quadratic $g(y) = 4y^2 + (4p+4)y + 1$ is a parabola opening upwards. The inequality $g(y) \ge 0$ holds for $y$ outside the roots of $g(y)=0$, if real roots exist. If $g(y)$ has no real roots, $g(y) > 0$ for all $y$, so the range is $(-\infty, \infty)$, which contains $[-1, -1/3]$. This case is not possible. $g(y)$ has real roots if the discriminant of $g(y)$ is non-negative. $\Delta_g = (4p+4)^2 - 4(4)(1) = 16(p+1)^2 - 16 = 16((p+1)^2 - 1)$. $\Delta_g \ge 0 \implies (p+1)^2 - 1 \ge 0 \implies (p+1)^2 \ge 1 \implies p+1 \ge 1$ or $p+1 \le -1 \implies p \ge 0$ or $p \le -2$.

If real roots $y_1, y_2$ exist ($y_1 \le y_2$), the range of $y$ where $g(y) \ge 0$ is $(-\infty, y_1] \cup [y_2, \infty)$. The condition that the range does not contain $[-1, -1/3]$ means that $[-1, -1/3]$ must be contained entirely within the region where $g(y) < 0$. Since $g(y)$ is a parabola opening upwards, $g(y) < 0$ between the roots $y_1$ and $y_2$. So, we must have $y_1 \le -1$ and $y_2 \ge -1/3$. The roots of $g(y)=0$ are $y = \frac{-(4p+4) \pm \sqrt{16((p+1)^2 - 1)}}{8} = \frac{-(p+1) \pm \sqrt{(p+1)^2 - 1}}{2}$. So, $y_1 = \frac{-(p+1) - \sqrt{(p+1)^2 - 1}}{2}$ and $y_2 = \frac{-(p+1) + \sqrt{(p+1)^2 - 1}}{2}$.

We need $y_1 \le -1$ and $y_2 \ge -1/3$. Condition 1: $y_1 \le -1$ $\frac{-(p+1) - \sqrt{(p+1)^2 - 1}}{2} \le -1$ $-(p+1) - \sqrt{(p+1)^2 - 1} \le -2$ $2 - (p+1) \le \sqrt{(p+1)^2 - 1}$ $1 - p \le \sqrt{(p+1)^2 - 1}$ This inequality holds if $1-p \le 0$ (i.e., $p \ge 1$) and $\Delta_g \ge 0$. If $p \ge 1$, then $p \ge 0$, so $\Delta_g \ge 0$ is satisfied. Thus, for $p \ge 1$, $y_1 \le -1$ holds. If $1-p > 0$ (i.e., $p < 1$), we must square both sides: $(1-p)^2 \le (p+1)^2 - 1$. $1 - 2p + p^2 \le p^2 + 2p + 1 - 1$ $1 - 2p \le 2p$ $1 \le 4p \implies p \ge 1/4$. Combined with $p < 1$, this gives $1/4 \le p < 1$. We also need $\Delta_g \ge 0$, which means $p \le -2$ or $p \ge 0$. For $1/4 \le p < 1$, we have $p \ge 0$, so $\Delta_g \ge 0$ is satisfied. The condition $y_1 \le -1$ holds for $p \ge 1$ or $1/4 \le p < 1$. Thus, $y_1 \le -1$ holds for $p \ge 1/4$ or $p \le -2$.

Condition 2: $y_2 \ge -1/3$ $\frac{-(p+1) + \sqrt{(p+1)^2 - 1}}{2} \ge -1/3$ $-(p+1) + \sqrt{(p+1)^2 - 1} \ge -2/3$ $\sqrt{(p+1)^2 - 1} \ge p+1 - 2/3$ $\sqrt{(p+1)^2 - 1} \ge p + 1/3$ We need $\Delta_g \ge 0$, so $p \le -2$ or $p \ge 0$. If $p+1/3 \le 0$ (i.e., $p \le -1/3$), the inequality $\sqrt{(p+1)^2 - 1} \ge \text{non-positive}$ is true whenever the square root is defined. The square root is defined for $p \le -2$ or $p \ge 0$. If $p \le -2$, then $p \le -1/3$ is true, so $y_2 \ge -1/3$ holds for $p \le -2$. If $p \ge 0$, we need to consider $p \le -1/3$. This is not possible for $p \ge 0$.

If $p+1/3 > 0$ (i.e., $p > -1/3$), we must square both sides: $(p+1)^2 - 1 \ge (p+1/3)^2$. $p^2 + 2p + 1 - 1 \ge p^2 + 2p/3 + 1/9$ $p^2 + 2p \ge p^2 + 2p/3 + 1/9$ $2p \ge 2p/3 + 1/9$ $4p/3 \ge 1/9$ $p \ge \frac{1}{9} \cdot \frac{3}{4} = \frac{1}{12}$. Combined with $p > -1/3$, this gives $p \ge 1/12$. We also need $\Delta_g \ge 0$, which means $p \le -2$ or $p \ge 0$. For $p \ge 1/12$, we have $p \ge 0$, so $\Delta_g \ge 0$ is satisfied. The condition $y_2 \ge -1/3$ holds for $p \le -2$ or $p \ge 1/12$.

We need both conditions to be satisfied: ($p \ge 1/4$ or $p \le -2$) AND ($p \ge 1/12$ or $p \le -2$). Let's analyze the intersection: If $p \le -2$, both conditions are satisfied. If $p > -2$: We need $p \ge 1/4$ and $p \ge 1/12$. The intersection is $p \ge 1/4$. So, the values of $p$ for which $y_1 \le -1$ and $y_2 \ge -1/3$ is $p \le -2$ or $p \ge 1/4$.

We also need to consider the case $y=0$. If $p \ne 0$, $y=0$ is in the range. The interval $[-1, -1/3]$ does not contain 0. So $y=0$ being in the range does not violate the condition. If $p=0$, the range is $(-\infty, 0) \cup (0, \infty) \setminus \{-1/2\}$. This range contains $[-1, -1/3]$. So $p=0$ is not allowed. The condition $p \le -2$ or $p \ge 1/4$ excludes $p=0$.

Thus, the true set of values of $p$ is $(-\infty, -2] \cup [1/4, \infty)$.