Answer the following questions

Question

f(x)= x^(1/3) * (x+8)

1. Domain of Def
2. Crit points and intervals increasing/ decreasing
3. Identify local max and min using first derivative test
4. Find absolute extrema

Answer 1

If you consider f(x)^3, as x^3 is a monotone transformation of x, what happens to this function, particularly at 0, becomes obvious. We see that this function will have a single minimum between -8 and 0, as the function has an odd number of roots at both points.

We could perhaps more easily solve this problem with x(x+8)^3, then transform via the cube root, but we will proceed directly.

1. The domain is all real numbers. If the denominator of a fractional power is odd, the domain is not restricted.
2. f'(x) = 1/3x^(-2/3)(x+8)+x^(1/3) = (1/3(x+8)+x)/x^(-2/3) = (4/3x + 8/3)/x^(2/3) = 4/3(x+2)/x^(2/3)

The derivative equals 0 at x = -2. The denominator equals 0 at 0, so 0 is a critical point also.

The function will be decreasing on (-∞, -2) and increasing on (2, ∞)

We note that, while there is a critical point at 0, the function is continuous there, and the derivative has a limit of ∞ on both sides of 0, so there will be an inflection at 0.

3 -2 is a local and global minimum. f(-2) = (-2)^(1/3)(6) = -6*2^(1/3) (The derivative is negative on x < -2 and positive on x > -2, so the function is convex there and this is a minimum.)

4 The absolute minimum is -6*2^(1/3) at x = -2. There is no absolute maximum as f(x) goes to infinity as x goes to ±∞