1. **Problem Statement:** Find the two critical values of the function $f(x) = 2x^3 - 18x^2 + 30x - 11$. 2. **Recall:** Critical values 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:** $$f'(x) = \frac{d}{dx}(2x^3 - 18x^2 + 30x - 11) = 6x^2 - 36x + 30$$ 4. **Set the derivative equal to zero to find critical points:** $$6x^2 - 36x + 30 = 0$$ 5. **Simplify by dividing both sides by 6:** $$x^2 - 6x + 5 = 0$$ 6. **Factor the quadratic:** $$(x - 5)(x - 1) = 0$$ 7. **Solve for $x$:** $$x = 5 \quad \text{or} \quad x = 1$$ 8. **Evaluate $f(x)$ at critical points to find critical values:** - For $x=1$: $$f(1) = 2(1)^3 - 18(1)^2 + 30(1) - 11 = 2 - 18 + 30 - 11 = 3$$ - For $x=5$: $$f(5) = 2(5)^3 - 18(5)^2 + 30(5) - 11 = 2(125) - 18(25) + 150 - 11 = 250 - 450 + 150 - 11 = -61$$ 9. **Conclusion:** The smaller critical value is $-61$ at $x=5$, and the larger critical value is $3$ at $x=1$. **Final answer:** Smaller critical value = $-61$, Larger critical value = $3$.