1. **State the problem:** Find the critical points of the function $f(x) = x^4 - 4x^3 + 4x^2 - 15$. 2. **Recall the formula:** Critical points occur where the derivative $f'(x)$ is zero or undefined. Since $f(x)$ is a polynomial, $f'(x)$ is defined everywhere, so we only solve $f'(x) = 0$. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(4x^3) + \frac{d}{dx}(4x^2) - \frac{d}{dx}(15) = 4x^3 - 12x^2 + 8x$$ 4. **Set the derivative equal to zero:** $$4x^3 - 12x^2 + 8x = 0$$ 5. **Factor out the greatest common factor:** $$4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 0$$ 6. **Solve each factor:** - From $4x = 0$, we get $x = 0$. - From $x^2 - 3x + 2 = 0$, factor further: $$x^2 - 3x + 2 = (x - 1)(x - 2) = 0$$ So, $x = 1$ or $x = 2$. 7. **List all critical points:** $$x = 0, 1, 2$$ **Final answer:** The critical points of $f(x)$ are at $x = 0$, $x = 1$, and $x = 2$.