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.
Please show me how to solve it.. is it using d/dx of u*v?
I still can't go beyond d/dx
Thanks.
1 Answer
Explanation:
We have:
f(x) = x(6-x)^(2/3)
An inflection point occurs when the second derivative vanishes.
If we differentiate wrt
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
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