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I have this inequality: -(x+5)(x+2)(1-x)(2x-8)<0. I want to find the solutions writing the "zeros" on the Real line and than write the function through the zeros I found. The signs of the function could be +,-,+,-,+ or -,+,-,+,- . Than i should chose the values "under" the x line (beacause the inequality is < 0 ).

What I'm asking is how do I chose to draw the function from "up" or from "down".

I'm a first year college student and I saw this method one time from in class. Please someone could explain me a little more?

Link for the ineqality and the plot

I have some more inequality to solve:

1- (x-2)(x-1)(4-x) >= 0
2- (x+3)(x-6)(x+5) >= 0
3- (x-1)(x+3)(x-5) >= 0
4- (x-2)(x-1)(4-x) >= 0
5- (x-3)(2-x)(x+7) >= 0

Explecially for the case 5, how the function could be "from the up" if i have a 3th grade?

All of those inequality are easy of course. But i need some explanation!

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Since the roots all have multiplicity $1$, there is a change of sign at every root. So if you want to know kmow to start, find which way the inequality goes somewhere. It is obvious that our function is positive for big $x$. Now the rest is easy to fill in. – André Nicolas Oct 8 '12 at 18:23
up vote 1 down vote accepted

I'll try to explain this.

As all the functions are continuous, all change of signs must occur only at the zeroes of the function,hence you mark all the zeroes in the real axis and you know that any change of sign can only occur at those zeroes due to continuity. Now Two conditions arise:

  1. The zero is non-repeating. It means that $f'(x) \neq 0$ at that point. Hence the function must change sign. To clarify it,consider that function doesn't changes sign ,then X-axis is a tangent and hence slope of tangent is $0$.

    So,you just plug one value (calculating sign of function at which is easier ) and find sign of function in that corresponding to that interval, then you just alternate the signs for all other intervals if the corresponding zeroes are non repeating.

  2. The zero is repeating. I am not very sure what will happen then. So I'll suggest you to find the values in next interval manually.

Now if you know the sign of function in each interval, to draw approximate graph,draw a free hand sketch. Note that this method does not tell you anything about values in intervals except whether they are positive or negative, so there you are on your own.


I'll try the first one: $-(x+5)(x+2)(1-x)(2x-8)=(x+5)(x+2)(x-1)(2x-8)$ with zeroes at $-5,-2,1,4$. Now, as $f(x) > 0$ for $x=100$ we just alternate signs of $f(x)$ at each zero. Hence $f(x) > 0$ in the intervals $(4, \infty ),(-2,1),(- \infty,-5)$ and $f(x) < 0$ for other intervals.

Note that there was nothing magical about $100$, you could use any number (though I think it would be better if you choose a number very loo or very high,i.e, corresponding to either $(- \infty ,a) $ or $(b,\infty )$. And then just alternate signs in intervals. Also you might want to watch out for repeated roots as I am not sure what to do. Just to be sure, check the next interval again.

In 5,check for value at say $-111$ it is positive. Now the function is positive in $(\infty, -7),(2,3)$ and negative elsewhere.

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Thank you, very exaustive. – Lc0rE Oct 8 '12 at 19:32

In the case of the inequality $-(x+5)(x+2)(1-x)(2x-8)<0$, you want to plot the graph of the curve $y = -(x+5)(x+2)(1-x)(2x-8).$ This has $x$-intercepts when $x=-5,-2,1$ or $4.$ Expanding the brackets gives $y = x^4 + \cdots$. This means it is the shape of a "wobbly smile". It is the same shape as $y = x^4$ but with extra "wobbles". The inequality is satisfied when the graph is below the $x$-axis, i.e. when $y < 0.$ This happens when $-5 < x < -2$ or $1 < x < 4$.

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Thanks for the answer. How about the other cases? I have a y = x^3 there... Expecially in the case 5... why it cames from the "up"? – Lc0rE Oct 8 '12 at 17:53
In the case $(x-3)(2-x)(x+7) \ge 0$ you need to plot the graph $y = (x-3)(2-x)(x+7).$ It crosses the $x$-axis when $x=-7,2$ or $3.$ Expanding the brackets gives $y = -x^3 + \cdots.$ The shape of the graph will be like $y = -x^3$, but with some more "wobbles". (When $x \ll 0$, $y \gg 0$ and when $x \gg 0,$ $y \ll 0.$) For $(x-3)(2-x)(x+7) \ge 0$ you want the graph $y = (x-3)(2-x)(x+7)$ to be above (or on) the $x$-axis. Thus $x \le -7$ or $2 \le x \le 3.$ – Fly by Night Oct 8 '12 at 19:04

Once you've found the points where each of these curves crosses the X-axis (the X-intercepts), you should pick one point within each of the segments between the X-intercepts, and also a point that's to the left of all of the X-intercepts, and a point that's to the right of all of the X-intercepts. Then just evaluate the function at each of those points to find out whether it's positive or negative. This will let you draw your curve the appropriate way around.

So, in your first example, you have X-intercepts at $x=-5,-2,1,4$, so you might evaluate the function for $x=-6,-3,0,2,5$ for example. This will clearly show you which way up the graph should be drawn.

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Thanks for the answer. Can you explain me "better" how to evaluate the function (-(x+5)(x+2)(1-x)(2x-8)<0) in x = −6,−3,0,2,5? – Lc0rE Oct 8 '12 at 18:07
I mean, for each of the chosen numbers in turn, replace $x$ in the expression with the number, and work it out. For example, using $x=-6$, you'd calculate $-((-6)+5)((-6)+2)(1-(-6))(2(-6)-8)$. – user22805 Oct 8 '12 at 18:11
Thanks. In the "case 3" how is why the function is "from up"? I tried to put x = -4 but nothing... – Lc0rE Oct 8 '12 at 18:18
I don't know what you're asking, sorry. – user22805 Oct 8 '12 at 19:15

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