Find the point of discontinuing of the given function: 2 Marks
f (x) = 3x + 1
X2-7x+12
I assume it is
f (x)
3x + 1
=--------
x²-7x+12
Factor the denominator into factors. Anytime the denominator becomes zero, there will be a discontinuity in the function, since the function at these points will be undefined (division by zero is not allowed).
The denominator factors to (x-3)(x-4) and therefore will have 2 positive roots.
Can you find them?
To find the point of discontinuity of the function, we need to determine the values of x for which the function is not defined.
The given function f(x) consists of two expressions: 3x + 1 and x^2 - 7x + 12.
The expression 3x + 1 is defined for all real numbers since it is a linear function.
Now let's analyze the expression x^2 - 7x + 12:
To determine the points of discontinuity, we need to find the values of x for which the expression x^2 - 7x + 12 is not defined.
The expression is a quadratic equation, and it is defined for all real numbers of x. Therefore, there are no points of discontinuity for the expression x^2 - 7x + 12.
Hence, the point of discontinuity for the given function f(x) = 3x + 1 and x^2 - 7x + 12 is none.
To find the points of discontinuity for the given function, we need to examine both functions individually.
1. Function f(x) = 3x + 1:
This is a linear function, which means it is continuous everywhere. Linear functions don't have any points of discontinuity.
2. Function g(x) = x^2 - 7x + 12:
To find the points of discontinuity for g(x), we need to look for values of x that make the function undefined or cause a vertical asymptote.
However, g(x) is a polynomial function, which is defined for all real numbers. There are no specific values of x that would cause the function to be undefined or create vertical asymptotes. Therefore, g(x) is continuous everywhere.
In summary:
- The function f(x) = 3x + 1 is continuous everywhere.
- The function g(x) = x^2 - 7x + 12 is also continuous everywhere.
Therefore, there are no points of discontinuity for the given functions.