Question

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

Answer 1

The curve is concave up when
`x<3`.

Explanation:

The given function is

`y=((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

`((f(x))/(g(x)))' =(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'=((x-3)(-1)-(1)(1-x))/((x-3)^(2) ) =(-x+3-1+x)/((x-3)^(2) ) =(2)/((x-3)^(2) )`

Then, we repeat the process to find the secon derivative

`y''=((x-3)^(2)(0)-2(x-3)(1)(2) )/((x-3)^(4) ) =(-4(x-3))/((x-3)^(4) )\\ y''=(-4)/((x-3)^(3) )`

For a concave up inflection, we must evalute
`y''>0`

`y''=(-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.

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