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*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.


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

enter image description here


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.


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 $3$: $[-\frac{13}4, 3)$.

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    $\begingroup$ it is not $|x-a|\lt x^2-3$, but $|x-a|\lt 3-x^2$ $\endgroup$ – farruhota Jun 4 '18 at 8:26
  • $\begingroup$ I feel I deserve upvotes (many). $\endgroup$ – jiten Jun 4 '18 at 11:48
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    $\begingroup$ +1, I never down-voted your post, I just indicated the mismatch. $\endgroup$ – farruhota Jun 4 '18 at 11:55
  • $\begingroup$ Request some analytical way, which is more succinct. Not happy with my approach. $\endgroup$ – jiten Jun 4 '18 at 12:12
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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 $$x^2+x-3<a<-x^2+x+3$$

enter image description here

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.

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  • $\begingroup$ Q.1. Seems like have forced the curve (parabola) on the negative x-axis. Cannot understand, as the parabola will be in both negative & positive x-axis, from $x=-\sqrt{3}$, to $x=\sqrt{3}$.The curve that should shift is $|x-a|$ by shifting the value of $a$ from negative value of $x$ to positive $x$. The left intersection should give maxima for the right arm of $x-a$ with the parabola, & vice-versa. Q.2. I have doubt at the end of my Updated post regarding the different method chosen by answer (at link) for finding maxima. Please help. Q.3. Please tell why taken $|a-x|$ for the inverted curve. $\endgroup$ – jiten Jun 4 '18 at 10:55
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    $\begingroup$ responded in the edit. $\endgroup$ – Siong Thye Goh Jun 5 '18 at 0:56
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    $\begingroup$ My approach and the linked solution is very similar. His solution is more rigorous than mine. farruhota's approach is brilliant in the sense he uses the slope. $\endgroup$ – Siong Thye Goh Jun 5 '18 at 5:59
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    $\begingroup$ to create a line that is parallel to $y$-axis, perhaps going through $(a,0)$, try $x=a$, you can then set the upper and lower limit. $\endgroup$ – Siong Thye Goh Jun 6 '18 at 0:50
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    $\begingroup$ change the $a$ in the constraint to $y$. $\endgroup$ – Siong Thye Goh Jun 6 '18 at 1:22
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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:

$\hspace{3cm}$enter image description here

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: $a\in (-3.25,3)$.

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  • $\begingroup$ Really good, and was thinking on that terms, but could not materialize the idea to use slope of $45^0$. $\endgroup$ – jiten Jun 4 '18 at 13:17
  • $\begingroup$ thank you and good luck! $\endgroup$ – farruhota Jun 4 '18 at 13:19
  • $\begingroup$ Please tell which s/w or online site is used to draw the displayed graph. It has plotted for different values of $a$ the function $|x-a|$. $\endgroup$ – jiten Jun 4 '18 at 13:24
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    $\begingroup$ It is MS Excel. I deliberately chose the values $a=-3.25; -2;0;3$ to draw, then showed the min/max values must be $-3.25$ and $3$. $\endgroup$ – farruhota Jun 4 '18 at 14:14

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