use the first derivate test to locate the maxima and minima of y = x^(1/3)
since y' = 1/3 * 1/x^(2/3) is never zero, there is no max/min.
Think of the graph. Cube root just gets bigger and bigger, the farther away from zero.
THANKS STEVE, YOU'RE THE BEST!
To apply the first derivative test to locate the maxima and minima of the function y = x^(1/3), we need to find the critical points first.
Step 1: Find the derivative of the function:
To find the derivative, we can apply the power rule, which states that if f(x) = x^n, then f'(x) = n * x^(n-1).
For y = x^(1/3), we can write it as y = x^(1/3) = x^(1/3) * 1.
Using the power rule, we differentiate to find y':
y' = (1/3) * x^(1/3 - 1) * 1 = (1/3) * x^(-2/3)
Step 2: Find the critical points:
To find the critical points, we set the first derivative equal to zero and solve for x:
(1/3) * x^(-2/3) = 0
Since there is no value of x that makes the first derivative equal to zero, there are no critical points.
Step 3: Determine intervals of increase and decrease:
To determine the intervals of increase and decrease, we look at the sign of the first derivative.
Since y' = (1/3) * x^(-2/3), the first derivative is positive (greater than zero) for all x values except x = 0 (since x^(-2/3) is undefined at x = 0).
Therefore, y is increasing for all x > 0 and decreasing for all x < 0.
Step 4: Determine the location of maxima and minima:
Since there are no critical points, we conclude that there are no local maximum or minimum points for the function y = x^(1/3).
However, we can see that the function has a global minimum at x = 0, where y = 0. This is because the function is defined as y = 0 for x = 0.
So, in summary, the function y = x^(1/3) has no local maxima or minima but has a global minimum at x = 0, y = 0.