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Math Challenge II-A Problem 2.18

 
 
JinTina的头像
Math Challenge II-A Problem 2.18
JinTina - 2022年04月3日 Sunday 15:07
 

Hello! I was reviewing this lecture, and I came to problem 2.18. It asks for how many values is f(x) negative. I got that x<-3 or x>2 b/c of (x+3)(x-2)<0. How is there 4 values? I looked at the solutions and it said something about roots negative in between something.  Can you help clarify? Thanks!


Tina 

 
LensmireJohn的头像
Re: Math Challenge II-A Problem 2.18
LensmireJohn - 2022年04月5日 Tuesday 11:07
 

Hi Tina,

There's two ways to understand the final answer here.

I would recommend a hybrid algebraic and graphical approach. Factoring like you did, you get that the graph of $y=x^2+x-6$ has zeros at $x=-3$ and $x=2$. Since the graph has a positive $x^2$ coefficient, it opens upward, so the graph looks like this (click on the graph below to view on Desmos):


This is where the "between" comes from. From the graph we can see it will be negative for $x$ values with $-3 < x < 2$. This gives the $4$ values $x=-2, -1, 0, 1$.

For the purely algebraic approach, factoring gives $(x+3)(x-2) < 0$. Here, for the product of two numbers to be negative, we need one to be positive and the other to be negative. Hence,

  • $x+3 < 0$ and $x-2 > 0$, implying that $x < -3$ and $x > 2$, but this is impossible.
  • $x+3 > 0$ and $x - 2 < 0$, implying $x > -3$ and $x < 2$, so $-3 < x < 2$ as before.

For the purely algebraic approach, you do need to be careful with the "and", as we did in the two cases above. Hope this helps!

JinTina的头像
Re: Math Challenge II-A Problem 2.18
JinTina - 2022年04月7日 Thursday 21:08
 

Thank you!