Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 3x^(1/3) + 6x^(4/3). Please use an analysis of f ′(x) and f ′′(x).

Answer 1

if y = 3x^(1/3) + 6x^(4/3)

then
dy/dx = x^-(2/3)+ 8 x^(1/3)

and
d^2y/dx^2
=-(2/3)x^-(5/3)+(8/3)x^-(2/3)

extrema or inflection where dy/dx = 0

note slope undefined when x = 0

dy/dx = x^-(2/3)+ 8 x^(1/3) = 0
or
1/x^(2/3) = - 8 x^(1/3)
1 = -8 x
x = -1/8

so what is d^2/dy^2 when x = -1/8 ?

d^2y/dx^2
=-(2/3)x^-(5/3)+(8/3)x^-(2/3)

= -(2/3)(-1/8)^-5/3 + (8/3) (-1/8)^-(2/3)

well -1/8^(5/3) = (-1/2)^5 = -1/32
so
(-1/8)^-5/3 = -32
and we have so far
-(2/3)(-32) +(8/3) (-1/8)^-(2/3)
64/3 + (-1/2)^-2 = 1/(1/4) = 4
64/3 + 4
so second derivative is positive, a relative minimum at x = -1/8
CHECK MY ARITHMETIC !!!

Answer 2

You got it right bro