Question

Y=log1,5(x^2-7x+12) найти область определения функции

Answer 1

$y=log_{1.5}(x^2-7x+12)$

В данном случае область определения будет будет: $x^2-7x+12\ \textgreater \ 0$
$D=7^2-4*12=49-48=1\\\\x_1=\frac{7+1}2=4\\\\x_2=\frac{7-1}2=3\\\\(x-3)(x-4)\ \textgreater \ 0$

Метод интервалов:

$........+.....3....-....4....+......$
$x\in (-\infty;3)U(4;+\infty)$

Answer 2

I can understand and communicate fluently in the user's language of choice such as English, 中文, 日本語, Español, Français, Deutsch, and others. I am chat mode of Microsoft Bing. I can help you with various tasks such as generating creative content, finding information, or solving problems.

To answer your question, the function $$Y=log_{1.5}(x^2-7x+12)$$ is defined only when the argument of the logarithm is positive, that is, when $$x^2-7x+12>0$$. To find the values of x that satisfy this inequality, we can factor the quadratic expression as $$(x-3)(x-4)>0$$. Then, we can use a sign chart to determine the intervals where the product is positive. The sign chart looks like this:

| x | -∞ | 3 | 4 | +∞ | |---|----|---|---|----| | (x-3) | - | - | + | + | | (x-4) | - | - | - | + | | (x-3)(x-4) | + | - | + | + |

From the sign chart, we can see that the product is positive when $$x<3$$ or $$x>4$$. Therefore, the domain of the function is $$(-\infty,3)\cup(4,\infty)$$. This means that the function is undefined when $$3\leq x\leq 4$$.

I hope this helps you understand the problem better. If you have any other questions, feel free to ask me. I am always happy to chat with you.Ответь+подробно.+Y=log1,5(x^2-7x+12)+найти+область+определения+функции