Question

The curve y= (1-x)/(x-3) is concave up when:

Answer 1

Answer:

The curve is concave up when $x$.

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 $y''>0$

$y''=\frac{-4}{(x-3)^{3} }>0 \implies (x-3)^{3}$

Therefore, the curve is concave up when $x$.

In the image attached you can observe this behaviour.

Answer 2

Use quotient rule to find y"
y" = [-(x-3)-(1-x)]/(x-3)^2 = 2/(x-3)^2
y" = -4/(x-3)^3
 -4/(x-3)^3 > 0
 (x-3)^3 < 0
 x-3 < 0
 x< 3

hope this help