1. **Problem:** Find the linear approximation of $f(x) = 3x e^{2x-10}$ at $x=5$. 2. **Formula:** The linear approximation (or tangent line approximation) 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 $f(x)$ evaluated at $x=a$. 3. **Step 1: Find $f(5)$** $$f(5) = 3 \times 5 \times e^{2 \times 5 - 10} = 15 e^{0} = 15 \times 1 = 15$$ 4. **Step 2: Find $f'(x)$** Use the product rule: if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. Here, $u(x) = 3x$ and $v(x) = e^{2x-10}$. - $u'(x) = 3$ - $v'(x) = e^{2x-10} \times 2$ (chain rule) So, $$f'(x) = 3 e^{2x-10} + 3x \times 2 e^{2x-10} = 3 e^{2x-10} + 6x e^{2x-10} = e^{2x-10}(3 + 6x)$$ 5. **Step 3: Evaluate $f'(5)$** $$f'(5) = e^{2 \times 5 - 10}(3 + 6 \times 5) = e^{0}(3 + 30) = 1 \times 33 = 33$$ 6. **Step 4: Write the linear approximation** $$L(x) = f(5) + f'(5)(x - 5) = 15 + 33(x - 5)$$ 7. **Final answer:** $$\boxed{L(x) = 15 + 33(x - 5)}$$