Question

What are the inflection points for f(x)= x (6-x) ^ 2/3 ?? 1. 3.6, 6 2. 3.6, 7.2 3. 6, 7.2 4. 3.6

Please show me how to solve it.. is it using d/dx of u*v?
I still can't go beyond d/dx
Thanks.

Answer 1

`x = 7.2`

Explanation:

We have:

`f(x) = x(6-x)^(2/3)`

An inflection point occurs when the second derivative vanishes.

If we differentiate wrt `x` and apply the product rule:

`d/dx (uv) = (u)(d/dx v) + (d/dx u)(v)`

along with the chain rule, and we get the first derivative:

`f'(x) = (x)(d/dx \ (6-x)^(2/3)) + (d/dx \ x)((6-x)^(2/3))`

`\ \ \ \ \ \ \ \ = x(2/3(6-x)^(-1/3)(-1)) + (1)((6-x)^(2/3))`

`\ \ \ \ \ \ \ \ = (6-x)^(2/3) - 2/3x(6-x)^(-1/3)`

`\ \ \ \ \ \ \ \ = 1/3(6-x)^(-1/3){3(6-x)^(2/3+1/3) - 2x }`

`\ \ \ \ \ \ \ \ = {3(6-x) - 2x } / (3(6-x)^(1/3))`

`\ \ \ \ \ \ \ \ = (18-5x) / (3(6-x)^(1/3))`

Then we differentiate again wrt `x` using the quotient rule:

`d/dx (u/v) = {v(d/dx u) - u(d/dx v)} / v^2`

to get the second derivative as follows:

`f''(x) = {3(6-x)^(1/3)(d/dx 18-5x) - (18-5x)(d/dx 3(6-x)^(1/3))} / (3(6-x)^(1/3))^2`

`\ \ \ \ \ \ \ \ \ \ = {3(6-x)^(1/3)(-5) - (18-5x)(3(1/3)(6-x)^(-2/3)(-1))} / (9(6-x)^(2/3))`

`\ \ \ \ \ \ \ \ \ \ = {(18-5x)(6-x)^(-2/3)-15(6-x)^(1/3)} / (9(6-x)^(2/3))`

`\ \ \ \ \ \ \ \ \ \ = (6-x)^(2/3)/(6-x)^(2/3) * {(18-5x)(6-x)^(-2/3)-15(6-x)^(1/3)} / (9(6-x)^(2/3))`

`\ \ \ \ \ \ \ \ \ \ = {(18-5x)(6-x)^(2/3-2/3)-15(6-x)^(2/3+1/3)} / (9(6-x)^(4/3))`

`\ \ \ \ \ \ \ \ \ \ = {(18-5x)-15(6-x)} / (9(6-x)^(4/3))`

`\ \ \ \ \ \ \ \ \ \ = {18-5x-90+15x} / (9(6-x)^(4/3))`

`\ \ \ \ \ \ \ \ \ \ = {10x-72} / (9(6-x)^(4/3))`

Then as per the earlier observation we have a point of inflection at coordinates where the second derivative vanishes, which requires that:

`f''(x) = 0`

`:. {10x-72} / (9(6-x)^(4/3)) = 0`

`:. 10x-72 = 0`

`:. x = 72/10`

`:. x = 7.2`