hint:Albert said:$x\in R \,\, and \,\, x\neq \pm 1$
$if$: $g(\dfrac {x-3}{x+1})+g(\dfrac {3+x}{1-x})=x$
$find$: $g(x)$
more hint :Albert said:hint:
let $g(\dfrac{x-3}{x+1})=a,g(\dfrac{x+3}{1-x})=b,g(x)=c$
find :$a,b,c$
so how to create 3 "Linear equations",and from them to get the solutions of $a,b,c$
The purpose of finding g(x) for x ∈ R and x ≠ ±1 is to determine the value of the function g(x) at any point on the real number line, except for x = ±1. This can help us understand the behavior of the function and make predictions about its values.
To calculate g(x) for x ∈ R and x ≠ ±1, we use the formula g(x) = (x²-1)/(x-1). This formula is derived from the definition of g(x) as the limit of (f(x)-f(1))/(x-1) as x approaches 1, where f(x) = x². By plugging in x ≠ ±1, we can find the value of g(x) at any point on the real number line.
The possible values of g(x) for x ∈ R and x ≠ ±1 are all real numbers, except for when x = ±1. This is because the function g(x) is defined as the limit of (f(x)-f(1))/(x-1) as x approaches 1, where f(x) = x². As long as x is not equal to ±1, the limit exists and the value of g(x) is a real number.
Yes, g(x) can be graphed for x ∈ R and x ≠ ±1. The graph of g(x) looks like a straight line with a break at x = 1, because the value of g(x) is undefined at this point. However, the graph is continuous and smooth for all other values of x.
For values of x close to ±1, g(x) behaves asymptotically. This means that as x approaches ±1, the value of g(x) approaches ±∞. This is because the denominator (x-1) approaches 0, causing the function to have a vertical asymptote at x = 1. This behavior is important to consider when analyzing the function and making predictions about its values.