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December 18th, 2017, 11:24 AM  #1 
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  How to find the correct statement in this cubic function?
$\textrm{The problem is as follows:}$ $\textrm{The function f(x) is defined as:}$ $$f(x)=x^{3}x+b$$ $\textrm{where xintercept is (a,0)}$ Which of the following statements for $a$ and $b$ are correct? $1.\,a<\frac{2\sqrt{3}}{3},\,b=a(1a^{2})$ $2.\,a>\frac{2\sqrt{3}}{3},\,b=a(a^{2}1)$ $3.\,a>\frac{9}{8},\,b=a(1a^{2})$ $4.\,a<\frac{9}{8},\,b=a(a^{2}1)$ $5.\,a>\frac{2\sqrt{3}}{3},\,b=a(1a^{2})$ What I tried to do was trying to plugin the $a$ value on the function given in order to find a relationship between $a$ and $b$. Since $(a,0)$ is the $xintercept$ Therefore, $$0=f(a)=a^{3}a+b$$ $$a^{3}a+b=0$$ $$b=aa^{3}=a(1a^{2})$$ By returning to the initial function I got to this: $f(x)=x^{3}x+(aa^3)$ $f(x)=x^{3}a^3x+a$ $f(x)=(xa)(x^2+ax+a^2)(xa)$ $f(x)=(xa)(x^2+ax+a^21)$ By finding the $xintercept$ where $f(x)=0$ both factors must equate to zero. Therefore by using the quadratic formula on the second term. $x_{1,2}=\frac{a\pm \sqrt{a^24(a^21)}}{2}$ The term in the square root must be non negative therefore; $a^24(a^21)\geqslant 0$ $3a^2\geqslant 4$ $a^2\leqslant \frac{4}{3}$ (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?. 
December 18th, 2017, 11:58 AM  #2 
Senior Member Joined: Dec 2015 From: holland Posts: 162 Thanks: 37 Math Focus: tetration 
$a \leqslant \sqrt{\frac{4}{3}} \frac{2\sqrt{3}}{3}\leqslant \ a\leqslant \frac{2\sqrt{3}}{3}$ Last edited by manus; December 18th, 2017 at 12:39 PM. 
December 18th, 2017, 12:15 PM  #3  
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  Quote:
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?.  
December 18th, 2017, 12:20 PM  #4 
Senior Member Joined: Dec 2015 From: holland Posts: 162 Thanks: 37 Math Focus: tetration 
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.

December 18th, 2017, 12:27 PM  #5  
Senior Member Joined: Dec 2015 From: holland Posts: 162 Thanks: 37 Math Focus: tetration  Quote:
 
December 18th, 2017, 12:36 PM  #6 
Senior Member Joined: Dec 2015 From: holland Posts: 162 Thanks: 37 Math Focus: tetration  
December 18th, 2017, 12:46 PM  #7 
Senior Member Joined: Dec 2015 From: holland Posts: 162 Thanks: 37 Math Focus: tetration  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.

December 18th, 2017, 02:13 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond 
$\sqrt{x^2}=x$

December 18th, 2017, 03:50 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,432 Thanks: 1462  
December 23rd, 2017, 01:57 PM  #10  
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  I was missing this identity. Thanks for clearing up that part!. Quote:
Quote:
I've posted a question above which I'm unsure. Can you help me to clear up my doubt?  

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