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Given function:
$f(x) = x^5 - 5x^4 + 5x^3 - 1$

Differentiating with respect to $x$:
$f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 1)(x - 3)$

For stationary/critical points, set $f'(x) = 0$:
$5x^2(x - 1)(x - 3) = 0 \implies x = 0, \; x = 1, \; x = 3$

Now, find the second derivative:
$f''(x) = 20x^3 - 60x^2 + 30x$

Testing each critical point:

1. At $x = 0$:
   $f''(0) = 0$
   Checking the sign of $f'(x)$ around $x = 0$: $f'(x) > 0$ for both $x < 0$ and $0 < x < 1$. Thus, $x = 0$ is a point of inflection, not an extremum.

2. At $x = 1$:
   $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 < 0$
   Hence, $f(x)$ has a local maximum at $x = 1$.
   $\text{Maximum value} = f(1) = 1^5 - 5(1)^4 + 5(1)^3 - 1 = 0$

3. At $x = 3$:
   $f''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$
   Hence, $f(x)$ has a local minimum at $x = 3$.
   $\text{Minimum value} = f(3) = 3^5 - 5(3)^4 + 5(3)^3 - 1 = 243 - 405 + 135 - 1 = -28$

Answer:

• Local Maximum Value: $0$ (at $x = 1$)
• Local Minimum Value: $-28$ (at $x = 3$)