Question

For f(x)= 3x+1 and g(x)= x^2-6, find (g/f)(x).

A. 3x+1/x^2-6 , x≠ ±√6

B. 3x+1/x^2-6

C. x^2-6/3x+1

D. x^2-6/3x+1, x≠ -1/3

Answer 1

D. The exclusion is important because we cannot divide by zero. Apart from that we are simply looking at a function. 

Answer 2

Answer:

$A. \frac{x^2+6}{3x+1}, x\neq -\frac{1}{3}$

Step-by-step explanation:

Given functions,

$f(x) = 3x + 1$

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

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

By substituting the values,

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

Which is a rational function,

We know that,

A rational function is defined for all real numbers except those for which denominator = 0,

If $3x+1 = 0$

$\implies 3x = -1$

$\implies x =-\frac{1}{3}$

i.e. domain restriction of g/f is x≠ -1/3

Hence, OPTION D is correct.