Answer:
- x = 0, local maximum
- x = ±(2/3)√3, global minima
Step-by-step explanation:
The first derivative is ...
 f'(x) = 30x^5 -40x^3 = 10x^3(3x^2 -4)
This will have zeros (critical points) at x=0 and x=±√(4/3).*
We don't need the second derivative to tell the nature of these critical points. Since the degree is even, the function is symmetrical about x=0. Since the leading coefficient is positive, it generally has a U-shape. This means the "outer" critical points will be minima, and the central one will be a local maximum.
__
However, since we're asked to use the 2nd derivative test first, we find the 2nd derivative to be ...
 f''(x) = 150x^4 -120x^2 = 30x^2(5x^2 -4)
For x=0, f''(0) = 0 -- as we expect for a function with a high multiplicity of the root at that point. For x either side of zero, both the function and the second derivative are negative, indicating downward concavity. That is, x = 0 is a local maximum.
For x² = 4/3, the second derivative is positive, indicating upward concavity. At x = ±√(4/3), we have local minima.
_____
* The "simplified" equivalent to √(4/3) is (2/3)√3.
