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