Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 174: 48

$$L(x)\approx -8x +48$$

Given $$ g(x)=\frac{x^{2}}{x-3}, \quad a=4$$ Since $g(4)=16$ and \begin{align*} g'(x)&=\frac{(x-3) 2 x-x^{2}(1 )}{(x-3)^{2}}\\ &=\frac{2 x^{2}-6 x-x^{2}}{(x-3)^{2}}\\ g'(4)&=-8 \end{align*} Then the linearization at $a=4$ is given by \begin{align*} L(x)&\approx g^{\prime}(a)(x-a)+f(a)\\ &=-8\left(x-4\right)+16\\ &=-8x +48 \end{align*}