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December 18th, 2017, 11:24 AM   #1
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Question How to find the correct statement in this cubic function?

$\textrm{The problem is as follows:}$

$\textrm{The function f(x) is defined as:}$

$$f(x)=x^{3}-x+b$$

$\textrm{where x-intercept is (a,0)}$

Which of the following statements for $a$ and $b$ are correct?

$1.\,|a|<\frac{2\sqrt{3}}{3},\,b=a(1-a^{2})$

$2.\,|a|>\frac{2\sqrt{3}}{3},\,b=a(a^{2}-1)$

$3.\,|a|>\frac{9}{8},\,b=a(1-a^{2})$

$4.\,|a|<\frac{9}{8},\,b=a(a^{2}-1)$

$5.\,|a|>\frac{2\sqrt{3}}{3},\,b=a(1-a^{2})$

What I tried to do was trying to plug-in the $a$ value on the function given in order to find a relationship between $a$ and $b$.

Since $(a,0)$ is the $x-intercept$

Therefore,

$$0=f(a)=a^{3}-a+b$$

$$a^{3}-a+b=0$$

$$b=a-a^{3}=a(1-a^{2})$$

By returning to the initial function I got to this:

$f(x)=x^{3}-x+(a-a^3)$

$f(x)=x^{3}-a^3-x+a$

$f(x)=(x-a)(x^2+ax+a^2)-(x-a)$

$f(x)=(x-a)(x^2+ax+a^2-1)$

By finding the $x-intercept$ where $f(x)=0$ both factors must equate to zero. Therefore by using the quadratic formula on the second term.

$x_{1,2}=\frac{-a\pm \sqrt{a^2-4(a^2-1)}}{2}$

The term in the square root must be non negative therefore;

$a^2-4(a^2-1)\geqslant 0$

$-3a^2\geqslant -4$

$a^2\leqslant \frac{4}{3}$ (not sure if this step is correct or could it be $a^2\geqslant \frac{-4}{-3}\geqslant \frac{4}{3}$?)

$a\leqslant \pm \sqrt{\frac{4}{3}}\leqslant \pm \frac{2\sqrt{3}}{3}$

Now this is where I'm confused as I do not know where does the absolute value comes from. Since solving a square root yields two values. I do not know which one should I choose. Can somebody help me with some theoretical background into this matter?.
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December 18th, 2017, 11:58 AM   #2
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$|a| \leqslant \sqrt{\frac{4}{3}} \frac{-2\sqrt{3}}{3}\leqslant \
a\leqslant \frac{2\sqrt{3}}{3}$

Last edited by manus; December 18th, 2017 at 12:39 PM.
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December 18th, 2017, 12:15 PM   #3
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Quote:
Originally Posted by manus View Post
$|a| \leqslant \sqrt{\frac{4}{3}} |a|\leqslant \frac{2\sqrt{3}}{3}$
Can you explain more about this part?. My question is about where does absolute value appears and more importantly why?

Why taking the square root of a makes the other side of the inequality only one value?. Please explain the theoretical base for this?. Could it be that you are taking the absolute value to both sides of the inequality?.
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December 18th, 2017, 12:20 PM   #4
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Because a is squared its graph is a parabola so if a^2 = x, a must be between - sqrtx and sqrtx, it is a parabola that opens upward.
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December 18th, 2017, 12:27 PM   #5
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Quote:
Originally Posted by Chemist116 View Post
Can you explain more about this part?. My question is about where does absolute value appears and more importantly why?

Why taking the square root of a makes the other side of the inequality only one value?. Please explain the theoretical base for this?. Could it be that you are taking the absolute value to both sides of the inequality?.
It is because a square is always positive so the quadratic equation has two solutions and a quadratic inequality has an interval as a solution (or two).
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December 18th, 2017, 12:36 PM   #6
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https://www.desmos.com/calculator/rgjxahaen6
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December 18th, 2017, 12:46 PM   #7
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Quote:
Originally Posted by Chemist116 View Post
Can you explain more about this part?. My question is about where does absolute value appears and more importantly why?

Why taking the square root of a makes the other side of the inequality only one value?.
With an inequality there exists a range of values, saying that the absolute value of x is less than a positive number a is the same as saying x is between -a and a.
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December 18th, 2017, 02:13 PM   #8
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$\sqrt{x^2}=|x|$
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December 18th, 2017, 03:50 PM   #9
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. . . $\textrm{where x-intercept is (a,0)}$
Is this meant to mean that $(a,\,0)$ is the only $x$-intercept?
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December 23rd, 2017, 01:57 PM   #10
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Quote:
Originally Posted by greg1313 View Post
$\sqrt{x^2}=|x|$
I was missing this identity. Thanks for clearing up that part!.

Quote:
Originally Posted by manus View Post
With an inequality there exists a range of values, saying that the absolute value of x is less than a positive number a is the same as saying x is between -a and a.
In the alternatives given all state as $|a|>\textrm{something}$ or $|a|<\textrm{something}$ However in my proposed solution I assumed $a^2-4(a^2-1)\geqslant 0$ and not $a^2-4(a^2-1)> 0$ This part I'm confused. Is there a justification or reason why to choose one over the other? At this point I'm changing my mind in favor to "make it like" the alternatives. But I would like to know the why?.

Quote:
Originally Posted by skipjack View Post
Is this meant to mean that $(a,\,0)$ is the only $x$-intercept?
The problem says so. So I'm assuming it is the only $x-intercept$.

I've posted a question above which I'm unsure. Can you help me to clear up my doubt?
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