Respuesta :
ANSWER
[tex]( \frac{f}{g} )(x) = \frac{4x + 1}{ {x}^{2} - 5 } .[/tex]
where,
[tex]x\ne \pm \sqrt{5} [/tex]
EXPLANATION
We were given that
[tex]f(x) = 4x + 1[/tex]
and
[tex]g(x) = {x}^{2} - 5[/tex]
We are supposed to find
[tex]( \frac{f}{g} )(x)[/tex]
Recall that,
[tex]( \frac{f}{g} )(x) = \frac{f(x)}{g(x)} [/tex]
We substitute the functions to obtain,
[tex]( \frac{f}{g} )(x) = \frac{4x + 1}{ {x}^{2} - 5 } [/tex]
This is a proper rational function and it is in the simplest form.
The restriction is that,
[tex]x\ne \pm \sqrt{5} [/tex]
If the denominator is zero the function will be undefined
[tex]( \frac{f}{g} )(x) = \frac{4x + 1}{ {x}^{2} - 5 } .[/tex]
where,
[tex]x\ne \pm \sqrt{5} [/tex]
EXPLANATION
We were given that
[tex]f(x) = 4x + 1[/tex]
and
[tex]g(x) = {x}^{2} - 5[/tex]
We are supposed to find
[tex]( \frac{f}{g} )(x)[/tex]
Recall that,
[tex]( \frac{f}{g} )(x) = \frac{f(x)}{g(x)} [/tex]
We substitute the functions to obtain,
[tex]( \frac{f}{g} )(x) = \frac{4x + 1}{ {x}^{2} - 5 } [/tex]
This is a proper rational function and it is in the simplest form.
The restriction is that,
[tex]x\ne \pm \sqrt{5} [/tex]
If the denominator is zero the function will be undefined
A function assigns the values. The value of (f/g)(x) is (4x+1)/(x²-5).
What is a Function?
A function assigns the value of each element of one set to the other specific element of another set.
Given the function f(x)=4x+1 and the function g(x)=x²-5, therefore, the value of (f/g)(x) will be,
(f/g)(x) = f(x)/g(x)
= (4x+1)/(x²-5)
Hence, the value of (f/g)(x) is (4x+1)/(x²-5).
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