Describe the x-values at which f is differentiable.
f(x)= x^2/5
I assume you mean x^(2/5)
since f' = 2/5 x^(-3/5)
it is undefined at x=0
To determine the x-values at which the function f(x) = x^2/5 is differentiable, we need to consider the differentiability of the function itself.
A function is differentiable at a particular point if the derivative exists at that point. The derivative of a function is the measure of its rate of change or slope at any given point.
In the case of f(x) = x^2/5, we can find the derivative using the power rule for differentiation. The power rule states that if we have a function of the form f(x) = x^n, the derivative is given by f'(x) = n * x^(n-1).
Applying the power rule to f(x) = x^2/5, we find:
f'(x) = (2/5) * x^(2/5 - 1)
= (2/5) * x^(-3/5)
Now, the function f(x) = x^2/5 is defined for all real values of x, so it exists everywhere. However, the derivative f'(x) = (2/5) * x^(-3/5) is only defined for x ≠ 0. This means that f(x) is differentiable for all x ≠ 0.
In conclusion, the x-values at which f(x) = x^2/5 is differentiable are all real numbers except x = 0.