1. **State the problem:** We are given the function $f(x) = 6\sqrt[3]{x}$ and asked to find the linearization $L(x)$ at $x=1$ and then use it to estimate $f(1.1)$. 2. **Recall the formula for linearization:** The linearization of a function $f(x)$ at $x=a$ is given by: $$L(x) = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the derivative of $f$ evaluated at $x=a$. 3. **Find $f(a)$:** Given $a=1$, $$f(1) = 6\sqrt[3]{1} = 6 \times 1 = 6$$ 4. **Find the derivative $f'(x)$:** Since $f(x) = 6x^{1/3}$, $$f'(x) = 6 \times \frac{1}{3} x^{-2/3} = 2x^{-2/3} = \frac{2}{x^{2/3}}$$ 5. **Evaluate $f'(a)$ at $x=1$:** $$f'(1) = \frac{2}{1^{2/3}} = 2$$ 6. **Write the linearization $L(x)$:** $$L(x) = f(1) + f'(1)(x - 1) = 6 + 2(x - 1)$$ 7. **Fill in the blanks:** The linearization is of the form $L(x) = [\text{coefficient}]x + [\text{constant}]$. Rewrite $L(x)$: $$L(x) = 6 + 2x - 2 = 2x + 4$$ So the blanks are: $$L(x) = 2x + 4$$ 8. **Use $L(x)$ to estimate $f(1.1)$:** $$f(1.1) \approx L(1.1) = 2(1.1) + 4 = 2.2 + 4 = 6.2$$ **Final answers:** - Linearization: $L(x) = 2x + 4$ - Estimate: $f(1.1) \approx 6.2$