Question

find the domain of function
f(x) = (x)^2+2x+3/x^2-5x+6, find the range of function​

Answer 1

Answer:

For this type of function, the domain is all real numbers. A function with a fraction with a variable in the denominator. To find the domain of this type of function, set the bottom equal to zero and exclude the x value you find when you solve the equation

Step-by-step explanation:

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.he domain of a function is the set of all acceptable input values (X-values). The range of a function is the set of all output values (Y-values).The domain of the expression is all real numbers except where the expression is undefined.

Answer 2

Answer:

Step-by-step explanation:

We have,

$y=f(x)=\frac{{x}^{2}+2x+3}{{x}^{2}-5x+6}$

$y=\frac{{x}^{2}+2x+3}{(x-3)(x-2)}$

This function to be defined, it's denominator shouldn't equals to 0.

so, the given function doesn't defined for x = 3 and x = 2

$dom(f) \in R - \{2 , 3\}$

Finding range:

$y = \frac{{x}^{2}+2x+3}{{x}^{2}-5x+6}$

$y{x}^{2} - 5yx + 6y = {x}^{2} + 2x + 3$

$(y - 1){x}^{2} - (5y + 2)x + (6y - 3) = 0$

For real values, it's discriminant must be greater than or equal to 0

So,

$(5y + 2)^{2} - 12(y - 1)(2y - 1) \geqslant0$

$25{y}^{2} + 20y + 4 - 12(2{y}^{2} - 3y +1) \geqslant 0$

${y}^{2} + 56y - 8\geqslant 0$

$(y + 28 - \sqrt{22})(y + 28 + \sqrt{22})\geqslant0$

$Range \in R - (-28+\sqrt{22}, -28-\sqrt{22})$