Question

Expressed powers are those that aref(x) = 5x + 3; g(x) = 6x - 5 Find f/g. (f/g)(x) = Quantity six x minus five divided by five x plus three. domain {x|x ≠ Five over six. } (f/g)(x) = Quantity five x plus three divided by six x minus five. domain {x|x ≠ Three over five } (f/g)(x) = Quantity five x plus three divided by six x minus five.domain {x|x ≠ Five over six. } (f/g)(x) = Quantity six x minus five divided by five x plus three. domain {x|x ≠ - Three over five. }

Answer 1

Answer:

(f/g)(x) = Quantity five x plus three divided by six x minus five.domain {x|x ≠ Five over six. }

Step-by-step explanation:

Given functions,

$f(x)=5x+3$

$g(x) =6x-5$

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

$=\frac{5x+3}{6x-5}$                        ( By substitution )

Let $h(x)=\frac{5x+3}{6x-5}$    

Which is the rational function,

Since, a rational function is defined for all real number except those for which denominator = 0,

$6x-5=0$

$6x=5$

$x=\frac{5}{6}$

Thus, h(x) is defined on all real numbers except 5/6,

Hence, domain of $\frac{f}{g}(x)$ is {x| x ≠ $\frac{5}{6}$ }

Answer 2

Answer: (f/g)(x) = Quantity five x plus three divided by six x minus five.domain {x|x ≠ Five over six. }

Explanation:

1) (f/g)(x) = f(x) / g(x)

2)

  f(x)         5x + 3
-------- = ------------
  g(x)        6x - 5

3) Since the division by 0 is not defined, the domain is restricted to 6x - 5 ≠ 0.

Therefore you must solve 6x - 5 = 0 to exclude the value of x for which the denominator becomes 0:

6 x - 5 = 0 => 6x = 5 => x = 5 / 6.

So, the domain is all x | x ≠ 5/6.