($a$ real number) So if $\frac{1}{4}<a<\frac{1}{3}$ prove that $\frac{10}9<R(a)<\frac{11}{6}$

where R(x)=$(2x-1)(x+1)(x-3)=2x^3-5x^2-4x+3$

So my idea was to do the same operations in $R(x)$ to $a$, I mean:

$$...<2a^3<...\\ ...<-5a^2<... \\ ...<-4x<... \\ ...<3<...$$ then I combine all of them. But this method takes lot of paper and a lot of time and I'm not sure if I will get a precise result. So is there a more efficient way to do it. Thank you in advance



If $p>0$ then from $\frac{1}{4}<a<\frac{1}{3}$ it follows directly that $\frac{p}{4}+q<pa+q<\frac{p}{3}+q$. You could apply that on the factors $2a-1$, $a+1$ and $a-3$. Based on these restrictions conclusions are possible (I hope, and didn't check that) for the product.

  • $\begingroup$ let me try, thanks $\endgroup$ – self Feb 25 '14 at 10:59
  • $\begingroup$ It requires less work but it works! :) thanks $\endgroup$ – self Feb 25 '14 at 11:14

$R(x)$ can be differentiated to yield the slope which is $R'(x) = 6x^2- 10x -4$. Notice that $x=2$ is a root of this quadratic. Let $R'(x)=k(x-a)(x-b)$ It is easy to see that $kab=-4$ where $k=6$, thus the other root is $\frac{-4}{ka}=\frac{-1}{3}$. Thus the derivative is negative between $x=-\frac{1}{3}$ and $x=2$. Thus the function is monotonically decreasing. You only need to plug in the values of $\frac{1}{4}$ and $\frac{1}{3}$ to find the upper and lower bounds respectively. Hope that helps. Have a graph of the function to help visualize.


Here's an alternate way to get the root, we'll call the derivative $g(x)$

$g(x) = 6x^2 - 10x - 4$

$g(x) = 6x^2 - 12 x + 2x - 4$

$g(x) = 6x(x-2) + 2(x-2)$

$g(x) = (x-2)(6x+2)$

Putting $g(x)=0$,

$(x-2) = 0 $ OR $(6x+2)=0$ [Either one of the factors must be zero]

$x=2$ OR $x=-\frac{1}{3}$


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