This problem is on my Calculus Readiness Test and I was having a lot of trouble with it. The problem is $$(x-3)(x-2)(x-1)gt0$$ I know how lớn solve $(x-3)gt0$ but I have never seen these type of problem before. I"ve tried lớn distribute everything but it gets messy và doesn"t really simplify neatly. How should I go about solving this?


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In fact you don’t want lớn multiply it out: the factored khung is much more useful here. You have a hàng hóa of three numbers; for the moment just hotline them $a,b$, and $c$. When is such a sản phẩm positive: it’s certainly positive if all three of $a,b$, và $c$ are positive. But it’s also positive if exactly two of them are negative & the remaining one is positive. There are the only ways lớn get a positive product from three factors.

Now, here’s a chart of the signs of the factors $x-3,x-2$, and $x-1$ & their product:

1 2 3 ----------------|------------|------------|-------------------- x-3: - - - - - 0 + x-2: - - - 0 + + + x-1: - 0 + + + + + product: - 0 + 0 - 0 +When $x3$. In interval notation, the solution set of the inequality is $(1,2)cup(x,infty)$.

This technique works whenever you’re comparing a product with $0$.

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edited Feb 27, 2013 at 14:03
answered Sep 8, 2012 at 2:17

Brian M. ScottBrian M. Scott
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As the others have mentioned, you must find the roots or zeroes of the function before solving the inequality. Therefore, the roots for $f(x) =(x-3)(x-2)(x-1)$ are $1,2,3$. There are numerous ways lớn solve this but I would lượt thích to show you how you can bởi vì it graphically. Here is the graph of the function:


Note, that the question asks you lớn find when $f(x) gt 0$ Therefore, look for all the spots on the graph where the graph is above zero and not equal to lớn zero. Firstly, we can disregard any value that is less than or equal lớn $1, 2, 3$ because at those points the function is not greater than zero.

Next, we can see from the graph that between the $1$ & $2$, the $f(x)$ is greater than zero. So we can say that one of the solutions is: $$ 1lt xlt2 $$ because at points between that interval, the value of $f(x) gt 0$.

Next, we disregard the interval $ 2 le x le 3$ because at that interval $x le 0$ or in other words, no matter what $f(x)$ you find between that interval, you will always have value that is $le 0$.

So, all we are left with is $xgt3$ because according lớn the graph, all values above $3$ are greater than $0$.

Therefore, the solution through graph analysis is $$1 lt x lt 2 \ x gt 3$$ or in interval notation $$ (1,2) cup (3,infty) $$