
Patrick B. answered 08/18/19
Math and computer tutor/teacher
y = x^(1/3)
y' = (1/3) x^(-2/3) = 1/(3*x^(2/3))
there are no values of x where the derivative is zero...
please repost with the correct function
Daria B.
asked 08/18/19True/False: "The only critical points of the function f(x) = x1/3 are x = 0 and x = ± 1/square root of 27"
Patrick B. answered 08/18/19
Math and computer tutor/teacher
y = x^(1/3)
y' = (1/3) x^(-2/3) = 1/(3*x^(2/3))
there are no values of x where the derivative is zero...
please repost with the correct function
Graph this function!
There are no maxima and minima.
There is a point of inflection at x=0.
Mark H. answered 08/18/19
Tutoring in Math and Science at all levels
I assume you mean
f(x) = x1/3
When I plot this, I see only one critical point. When x = 0, f(x) = 0, and there is a local positive slope. For x values less than ±1, the absolute value of f(x) is greater than x, and for larger values of x, f(x) is smaller.
Nothing special happens at x = 27
Always plot thing like this to help visualize. Here is my favorite online plotting tool:
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