# Question about method in an algebra precalculus exercise

($a$ 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

Hint:

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.

• let me try, thanks – self Feb 25 '14 at 10:59
• It requires less work but it works! :) thanks – 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 $$x=-\frac{1}{3}$$