Find real $a$ for which $x^2 +|x-a|-3\lt 0$ is satisfied for negative $x$.

Question

*Have problem solution given at: https://math.stackexchange.com/a/460988/424260, which am unable to understand. *

This problem's graphical soln. takes $|x-a|\lt 3-x^2$, & has right arm of the graph for $|x-a|$ first touching the curve for negative $a$, & then the left arm for positive $a$, with range between these two values of $a$ being asked for.

Am unable to get the stated values as given at the above link's answer :$$x^2+x-3\lt a\lt -x^2+x+3$$ has both l.h.s. ($x^2+x-3$) & r.h.s. ($-x^2+x+3$) having the roots given by same values $x= \frac{-1\pm \sqrt{13}}{2}$.

So, the second para. is incomprehensible to me.

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Update The approach chosen by the solution's analytical approach (also the one desired by me) is to not consider the two graphs separately, but to consider the points for which the modulus changes from $(x-a) \lt 0 $ to $\ge 0$. The range gives the value of $a$, & need take minima of the left side $(x^2+x-3)$ that represents the intersection of the right arm of the term $|x-a|$ with the parabola, as shown below:
$x-a\ge 0\implies x^2+x-3\lt a$

So, the minima represents the minimum positive value of $a$ that would yield the intersection of the right arm of $|x-a|$ with the parabola. This is minima as it is the smallest value of $x$ for the 'combined' curve, where any intersection is possible.

This minima can be found by taking derivative of the equality : $x^2+x-3=a$ wrt $x$, i.e. $\frac{\delta a}{\delta x} = 2x +1$, and taking second dervative to verify minima, get : $\frac{\delta^2 a}{\delta^2 x} = 2$; a positive value; hence minima. This minima is given at : $x = -\frac 12$. For this value of $x$, the value of $x^2 +x-3 = \frac14-\frac 12-3= -\frac {13}{4}$.

Next, need find maxima , as the left arm would intersect the curve; leading to 'cumulative' curve having a maximum value of $x$.

This maxima can be found by taking derivative of the equality : $-x^2+x+3=a$ wrt $x$, i.e. $\frac{\delta a}{\delta x} = -2x +1$, and taking second dervative to verify maxima, get : $\frac{\delta^2 a}{\delta^2 x} = -2$; a negative value; hence maxima. This maxima is given at : $x = \frac 12$. For this value of $x$, the value of $-x^2 +x-3 = -\frac14+\frac 12+3= \frac14 +3 = \frac{13}{4}$.

Have two issues here:
(i) as the solution (at the given link) takes the same value of minima $x=-\frac12$ to get the maxima point for intersection of the left arm for positive value of $x$, by substituting in $-x^2+x+3$, i.e. $-\frac14-\frac12+3\implies -\frac34 +3\implies \frac94$.

(ii) the value obtained by me for maxima is invalid as it should be less than $\sqrt{3}$; but it is greater than that.
However, the values obtained by me for maxima ($\frac{13}4 = 3.25$) and minima (-$\frac{13}4 = -3.25$) values of $a$ in terms of $x$ are correct graphically, as shown at: https://www.desmos.com/calculator/h0fuoqe7jc

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Update 2 The link has given a very nice idea to just take the minima instead of finding roots of the concerned two quadratic equations; but it seemingly falters in finding maxima & it uses the symmetric properties of both hyperbola & $|x-a|$ to unnecessarily find the intersection of left arm too at the same $x$ value.

The parabola is symmetrical around the y-axis, hence it is logical that $a$ values (for maxima & minima) too are just a sign change only. Similarly, the function $|x-a|$ too is. In fact symmetric property is used by solution (at the link given) to derive the left arm 's value for $a$ for given $x=\frac12$.
In fact the value $a=\frac94$ is not apparent as to where it should belong to, as it gives two intersections of the left arm with parabola. Am very confused at its graphical interpretation.
The only possible interpretation is that the second intersection point is having coordinates $x=-\frac12, y = \frac{11}4$, i.e. the chosen value of minima ($x=-\frac12$). So, it was a wrong idea for the solution (at given link) to put the minima value for left arm equation to find maxima.

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Update 3 Need find not maxima, but the value of $a$ for which negative $x$ values are there. So, from minima $-\frac{13}4$ to $x\lt 3$ for which the left arm intersects parabola at y-axis. Will arrive at the value $a=3$ as below:
The $y$ coordinate is same for both parabola and the left arm intersection at y axis. It is a unique point. The value $-x^2+x+3= a$ gives the left arm equation. But, $x=0$, so $a=3$.
So, the value range for $a$ is from minima to less than $3$: $[-\frac{13}4, 3)$.

Answer 1

Preparation:

• If we encounter an inequality $|x|<\beta$. If $\beta \le 0$, then no such $x$ exists. Otherwise, it is equivalent to $-\beta < x< \beta$.

Now, when we encounter the inequality $x^2+|x-a|-3<0$, this is equivalent to $$|x-a| < 3-x^2.$$

$$|a-x|< 3-x^2$$

Hence to have a solution, we need the condition that $3-x^2>0$, that is we are interested in the solution in $(-\sqrt3,0)$. Also, from the preparation material, we let $\beta=3-x^2$, and we have

$$-(3-x^2)<a-x < 3-x^2$$

which is just $$x^2+x-3<a<-x^2+x+3$$

Hence $a$ take values from $(-\frac{13}4, 3).$

Remark: It is puzzling why you write $|x-a|<x^2-3$.

Edit $1$:

• The question of interest is find the value of $a$ when a negative $x$ exists that satisfies the inequality. Hence, we just have to look at negative part. In fact, we just have to look at $(-\sqrt3, 0)$. In fact, it is within this range that we have $x<0$ and also $x^2+x-3 < -x^2+x+3$ which is equivalent to $2(x^2-3)<0$.

• The curve $x^2+x-3=\left( x+\frac12\right)^2-3-\frac14=\left( x+\frac12\right)^2-\frac{13}4$ attains the minimal value at point $\left(-\frac12, -\frac{13}4 \right)$. Note that we have $-\sqrt{3} < -\frac12$.

• The curve $-x^2+x+3=-(x^2-x)+3=-\left(x-\frac12 \right)^2+3+\frac14$ which attains maximal value at point $\left(\frac12, \frac{13}4 \right)$. The maximal is attained at point $\left( \frac12, \frac{13}4\right)$. However, recall that we are interested in $-\sqrt{3}<x<0$, hence the supremum is $-0^2+0+3=3$.

• Hence $-\frac{13}4<a<3$

• Here is a desmos link, we can see when does $y=a$ lies in the region of interest.

Edit $2$:

The set of interest is \begin{align}A&=\{a: \exists x \in (-\sqrt{3}, 0), y_1(x) < a <y_2(x)\} \\ &= \{a: \exists x \in (-\sqrt{3}, 0), y_1(x) < y <y_2(x), y=a\} \\\end{align}

Let $D=\{(x,y): -\sqrt{3}<x<0, y_1(x) < y<y_2(x) \}$, we are interested in knowing for which value of $a$ does the $y=a$ intersect with $D$.

Also notice that $D$ doesn't include it's boundary. Hence $a\ne -\frac{13}4$.

Here is another desmos link that shaded the region in $D$. notice that $D$ doesn't include any boundary point. We can see that $A=(\inf_{(x,y) \in D}y, \sup_{(x,y)\in D}y)=\left( - \frac{13}4, 3 \right).$

Remark: The original solution in the previous posts is more rigorous. My approach uses a graph to help visualization.

Answer 2

Determine all the values of the parameter $a$ for which the inequality $3−|x−a|>x^2$ is satisfied by at least one negative $x$.

Graphical method: at least one negative solution exists for $|x-a|\lt 3-x^2$ implies that the graph of absolute value function lies below the graph of quadratic function for $x<0 \ $ ($-\sqrt{3}<x<0 \ $ to be precise).

Refer to the graph below:

$\hspace{3cm}$

The values of $a$ must range between $a_1$ and $a_2$ for the hand of absolute value function to stay below the parabola for $x<0 \ $ ($-\sqrt{3}<x<0 \ $ to be precise).

Since the slope of the hand is $1$, it is easy to find $a_2=3$ from $B(0,3)$.

The point $A$ is a tangent point to the parabola of a tangent line with slope $1$. Hence: $$y'=(3-x^2)'=-2x=1 \Rightarrow x_0=-0.5 \Rightarrow y_0=3-(-0.5)^2=2.75.$$ The tangent line is: $$y=2.75+1\cdot (x-(-0.5))=x+3.25,$$ whope $x$-intercept is $(-3.25,0)$, hence: $a_1=-3.25$.

At last, the solution is: $a\in (-3.25,3)$.