find the absolute minimum and maximum values of f(x)=x^(1/3) (3+x) on the interval [-2,1]

f' = (4x+3)/(3x^(2/3))

This is undefined at x=0, where f(0) = 0, a local min.

f'=0 at x = -3/4 where there is a local max. f(-3/4) = -(9/4) ∛(3/4)

check the value of f at the end points for other max/min values.

do not get it :(

what don't you get?

max/min occurs where f' = 0, as long as f" is not also zero

where f' is undefined, check to see whether f is defined there. If so, it will be some kind of cusp, so a max or min. Or it will have a vertical tangent.

As it turns out, I made a mistake at x=0; there is no local min there.
And, the extremum at x = -3/4 is a min, not a max.

Oddly, this error was caused by my trust in the graphs at wolframalpha.com, which showed it wrong.

To see the real graph, visit

http://rechneronline.de/function-graphs/

To find the absolute minimum and maximum values of the function f(x) = x^(1/3) (3+x) on the interval [-2, 1], we need to follow these steps:

1. Find the critical points: These are the points where the derivative of the function is either zero or undefined.

2. Identify the endpoints: The function needs to be evaluated at the endpoints of the interval [-2, 1].

3. Evaluate the function at the critical points and endpoints: Substitute the critical points and endpoints into the function and find the corresponding function values.

4. Compare the function values: Compare all the function values obtained from step 3 to determine the absolute minimum and maximum values.

Let's go through these steps in detail:

Step 1: Find the critical points
To find the critical points, we need to find where the derivative of f(x) equals zero or is undefined.

First, let's find the derivative of f(x):
f'(x) = (1/3) x^(-2/3) (3+x) + x^(1/3) (d/dx)(3+x)

Simplifying the derivative:
f'(x) = (1/3) x^(-2/3) (3+x) + x^(1/3)

To find the critical points, set f'(x) equal to zero and solve for x:
(1/3) x^(-2/3) (3+x) + x^(1/3) = 0

Multiplying through by 3x^(2/3) to eliminate the denominators:
(3+x) + 3x = 0

Simplifying:
6x + 3 = 0
6x = -3
x = -1/2

So, the critical point is x = -1/2.

Step 2: Identify the endpoints
The interval is [-2, 1], so the endpoints are -2 and 1.

Step 3: Evaluate the function at the critical point and endpoints
Substitute the critical point and endpoints into the original function f(x) = x^(1/3) (3+x) to find the corresponding function values:

f(-2) = (-2)^(1/3) (3+(-2)) = (-2)^(1/3) (1) = -2^(1/3)
f(-1/2) = (-1/2)^(1/3) (3+(-1/2)) = (-1/2)^(1/3) (5/2)
f(1) = (1)^(1/3) (3+1) = (1)^(1/3) (4)

Step 4: Compare the function values
To identify the absolute minimum and maximum values, compare the function values obtained in step 3.

f(-2) = -2^(1/3) ≈ -1.587
f(-1/2) ≈ (-1/2)^(1/3) (5/2) ≈ 1.48
f(1) = (1)^(1/3) (4) = 4

From the comparison, we can see that the absolute minimum value of f(x) is -1.587 at x = -2, and the absolute maximum value is 4 at x = 1.