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Using the concept of domain, the domain of (f.g)(x) is given by:
{x ∈ ℝ | x ≠ 3}
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- The domain of a function is given by all possible input values, that is, <u>on a graph, all values that the x-axis assumes.</u>
- In the graph, <u>function f assumes all real values.</u>
- Function g is not defined for x = 3, thus, it's domain is all real values except 3.
- Thus, the multiplication, as
, will also not be defined at x = 3, and the domain of the multiplication is:
{x ∈ ℝ | x ≠ 3}
A similar problem is given at brainly.com/question/4175434
A discontinuity is a point that cannot exist because the x-coordinate would cause a problem in the equation. If you have a polynomial in the denominator, you must find which values of x would cause the polynomial in the denominator to evaluate to zero. Since division by zero is undefined, that would cause a discontinuity.
Let's look at your function.
is in the numerator. It is defined for every value of x. There is no problem there.
is in the denominator. This is a function defined for every value of x, but since it is in the denominator, we must exclude the x-value that would cause this polynomial to evaluate to zero.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is:
Let's look at your function.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is: