question 7 (1 point)\nwhere is the function ( f(x)=\frac{x - 4}{2x - 6} ) increasing?\n( \bigcirc ) a) ( x…

question 7 (1 point)\nwhere is the function ( f(x)=\frac{x - 4}{2x - 6} ) increasing?\n( \bigcirc ) a) ( x geq 4, x lt 3 )\n( \bigcirc ) b) ( x in mathbb{r}, x \neq 3 )\n( \bigcirc ) c) ( x gt 6 )\n( \bigcirc ) d) ( x gt 3 )

question 7 (1 point)\nwhere is the function ( f(x)=\frac{x - 4}{2x - 6} ) increasing?\n( \bigcirc ) a) ( x geq 4, x lt 3 )\n( \bigcirc ) b) ( x in mathbb{r}, x \neq 3 )\n( \bigcirc ) c) ( x gt 6 )\n( \bigcirc ) d) ( x gt 3 )

Answer

Explanation:

Step1: Find the derivative of (f(x))

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}). Here (u = x - 4), (u^\prime=1), (v = 2x - 6), (v^\prime = 2). [ \begin{align*} f^\prime(x)&=\frac{1\times(2x - 6)-(x - 4)\times2}{(2x - 6)^2}\ &=\frac{2x-6-(2x - 8)}{(2x - 6)^2}\ &=\frac{2x-6 - 2x + 8}{(2x - 6)^2}\ &=\frac{2}{(2x - 6)^2} \end{align*} ]

Step2: Determine where (f^\prime(x)>0)

Since ((2x - 6)^2>0) for all (x\neq3) (because the square of a non - zero real number is positive) and the numerator (2>0). So (f^\prime(x)>0) for all (x\in R,x\neq3)

Answer:

B. (x\in R,x\neq3)