Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 62

$$L(x) = -5x +12$$

Given $$f(x) =x e^{6-3 x} ,\ \ a=2$$ Since $f(a) =2$ and \begin{align*} f'(x)&= \frac{d}{dx}\left(x\right)e^{6-3x}+\frac{d}{dx}\left(e^{6-3x}\right)x\\ &= (1-3x ) e^{6-3 x}\\ f'(2)&= -5 \end{align*} Then the linearization is given by \begin{align*} L(x)&=f^{\prime}(a)(x-a)+f(a)\\ &= -5 (x-2)+2\\ &= -5x +12 \end{align*}