I want to find the invariant points( points that are fixed) algrebraically of x^2 - 25 and its reciprocal. Would I set up my equation like this?
0 = x^2 -25
or
0 = 1/ x^2 - 25
Thanks
Knowing the definition of specific math terms helps solving the problem, most of the time.
Here the term is invariant point, which is usually called fixed point, defined as follows (Wiki)
"In mathematics, a fixed point (sometimes shortened to fixpoint, also known as an invariant point) of a function is an element of the function's domain that is mapped to itself by the function. That is to say, c is a fixed point of the function f(x) if and only if f(c) = c."
So invariant points can be solved by equating f(x)=x, in the given case,
x²-25=x
x²-x-25=0
Solve for x using the quadratic formula.
To find the invariant points algebraically, we need to set the equations equal to zero and then solve for x. However, in order to find the invariant points of the reciprocal function, we first need to find the invariant points of the function itself.
For the equation x^2 - 25 = 0, you set the equation equal to zero because we are looking for the values of x where the function is equal to zero. By solving this equation, we can find the invariant points of the function x^2 - 25.
Let's solve the equation x^2 - 25 = 0 using algebraic methods:
1. Add 25 to both sides of the equation:
x^2 - 25 + 25 = 0 + 25
x^2 = 25
2. Take the square root of both sides (remembering to consider both positive and negative solutions):
x = ±√25
The square root of 25 is 5, so we have:
x = ±5
Hence, the invariant points of the function x^2 - 25 are x = -5 and x = 5.
Now, let's find the invariant points of its reciprocal function. The reciprocal function is obtained by taking the reciprocal of each invariant point of the original function.
For the equation 1/(x^2 - 25) = 0, you do not set the equation equal to zero. Instead, we will use the invariant points of the original function x^2 - 25, which are x = -5 and x = 5, and find their reciprocals to find the invariant points of the reciprocal function.
1. Reciprocal of -5:
1/(-5) = -1/5
2. Reciprocal of 5:
1/5
Therefore, the invariant points of the reciprocal function 1/(x^2 - 25) are x = -1/5 and x = 1/5.
In summary:
- The invariant points of the function x^2 - 25 are x = -5 and x = 5.
- The invariant points of the reciprocal function 1/(x^2 - 25) are x = -1/5 and x = 1/5.