i was doing some pre-calc until i stumbled on this problem:
$$(x^2-2x-3)^{1999}(x^2-6x+5)^{2000}\ge0$$
Powering to 1999 is like powering to 1, so i wrote:
$$(x^2-2x-3)^{1999}\ge0 = (x^2-2x-3)\ge0$$
Solved it, and the answers are -1,3. When i study the sign, i get $1\le x \land x\ge 3$
After i graph it, i go to the main problem:
$$(x^2-6x+5)^{2000}\ge0$$
Powering to 2000 is like powering to $2$, so i wrote
$$(x^2-6x+5)^{2000}\ge0 = (x^2-6x+5)^2\ge0$$
So my thoughts are:
- I know that any number powered to 2 in this case will be bigger or equal to 0. After i solve it (roots are 5,1), how should i approach?
- At the end, the answer is $(-\infty,-1]\cup(1)\cup[3,+ \infty)$
Why should i put exclude the one?
And what would happen if the inequality would be reversed?
The term $(x^2-6x+5)^2$ doesn't change the sign when it is positive and make the inequality satisfy at equality when it is equal to zero.
From solving $x^2-2x-3 \geq 0$, you have obtain $x \le -1$ or $x \ge 3$.
Solving $x^2-6x+5=0$, $(x-1)(x-5)=0$ means we have to include $1$ and $5$ into our solution set and hence, the result.
In summary:
$$(x^2-2x-3)^{1999}(x^2-6x+5)^{2000}\ge0$$
is equivalent to -
$$x^2-2x-3 \geq 0 \text{ or } x^2-6x+5=0$$.