Find the intervals of concavity and inflection points of f(x) = 5x 2/3 − 2x 5/3

Answer 1

Assuming you mean f(x) = 5x^{2/3 -} 2x^5/3

Note the function is zero at x = 0 and complex if x < 0. focusing on the real part,

The first derivative is f'(x) = 10 ⁄ 3x^1/3 - 10x^2/3 / 3

and the second derivative is f''(x) = -20 / 9x^1/3 - 10 / 9x^4/3

If you set f'(x) = 0, you get 0 = 10 ⁄ 3x^1/3 - 10x^2/3 / 3 => 10/3 * (1/x^1/3 - x^2/3/1) = 0 (1)

since x^2/3/1 * x^1/3/x^1/3 = x/x^1/3, we can solve (1) to be

10/3 * (1-x) / x^1/3 = 0

If you use x = 1, then the equation is true since 1-1 = 0. There is a critical point at x = 1.

Plug x = 1 into f''(x) and you get f''(1) = -20/9 - 10 /9 which is negative. so that the function has a maximum at this point.

Since there is only one critical point at x=1 and it is a maximum, we can say the the entire function is concave with interval [0,1] ∪ [1,∞).

You can also plug values of x into the original f(x) and f'(x) to check

f(x) at x =1 is 3 so the maximum is(1,3)

f'(1/2) = 2.099

f'(1) = 0

f'(1.5) = -1.4

f'(10) = - 13.9

ie, as it goes from 0 to 1, it is increasing, as it goes from 1 to infinity it is decreasing, showing concave with inflection point at x = 1.

Also note that f'(0) is undefined since there is an asymptote at this point in the derivative. Cant divide by zero. Or another way to say it is as x→0, f'(x) →∞.