Answer:
Step-by-step explanation:
The given function is
[tex]y=\frac{(1-x)}{(x-3)}[/tex]
To find where the function is concave up, we have to find the second derivative of the function.
The derivative of a fraction is
[tex](\frac{f(x)}{g(x)})' =\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^{2} }[/tex]
Where [tex]f(x)=1-x[/tex] and [tex]g(x)=x-3[/tex].
Using the property, we have
[tex]y'=\frac{(x-3)(-1)-(1)(1-x)}{(x-3)^{2} } =\frac{-x+3-1+x}{(x-3)^{2} } =\frac{2}{(x-3)^{2} }[/tex]
Then, we repeat the process to find the secon derivative
[tex]y''=\frac{(x-3)^{2}(0)-2(x-3)(1)(2) }{(x-3)^{4} } =\frac{-4(x-3)}{(x-3)^{4} }\\ y''=\frac{-4}{(x-3)^{3} }[/tex]
For a concave up inflection, we must evalute [tex]y''>0[/tex]
[tex]y''=\frac{-4}{(x-3)^{3} }>0 \implies (x-3)^{3} <0 \implies x-3<0 \implies x<3[/tex]
Therefore, the curve is concave up when [tex]x<3[/tex].
In the image attached you can observe this behaviour.
