Question

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

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 \(u = x - 4\), \(u^\prime=1\), \(v = 2x - 6\), \(v^\prime = 2\).

$$
LATEXBLOCK0
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Step2: Determine where \(f^\prime(x)>0\)

Since \((2x - 6)^2>0\) for all \(x
eq3\) (because the square of a non - zero real number is positive) and the numerator \(2>0\). So \(f^\prime(x)>0\) for all \(x\in R,x
eq3\)

Answer:

B. \(x\in R,x
eq3\)