Question

Pleaseeeee help asap!!

Answer 1

ANSWER

$D. \: \: \frac{ {x}^{2} - 6}{3x + 1} \: ,x \ne - \frac{1}{3}$



EXPLANATION

The given functions are;


$f(x) = 3x + 1$


and

$g(x) = {x}^{2} - 6$


We want to find

$( \frac{g}{f} )(x)$


Recall that,


$( \frac{g}{f} )(x) = \frac{g(x)}{f(x)}$


This implies that,

$( \frac{g}{f} )(x) = \frac{ {x}^{2} - 6}{3x + 1}$

The restriction on this function is that, the denominator must not be zero.


Thus

$3x + 1 \ne0$



$\Rightarrow x \ne - \frac{1}{3}$

Answer 2

Answer:

(D) $(\frac{g}{f})(x)=\frac{x^{2}-6}{3x+1}, x{\neq}-\frac{1}{3}$

Step-by-step explanation:

Given: f(x)=$3x+1$ and g(x)=$x^{2}-6$.

To find: $(\frac{g}{f})(x)$

Solution: Given that f(x)=$3x+1$ and g(x)=$x^{2}-6$, then applying the operations on the functions f(x) and g(x), we get

$(\frac{g}{f})(x)= \frac{g(x)}{f(x)}$, f(x)≠0

$(\frac{g}{f})(x)=\frac{x^{2}-6}{3x+1}$

Thus, $(\frac{g}{f})(x)=\frac{x^{2}-6}{3x+1}, x{\neq}-\frac{1}{3}$

Hence, option D is correct.