if (f(x)=4x+\frac{5}{x}), find (f(-3)), using the definition of derivative. (f(-3)) is the limit as…

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function $y = f(x)$ is $f^{\prime}(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}$. Here $f(x)=4x+\frac{5}{x}$ and $a=-3$, so $f(-3)=4\times(-3)+\frac{5}{-3}=-12-\frac{5}{3}=-\frac{36 + 5}{3}=-\frac{41}{3}$. Then $\frac{f(x)-f(-3)}{x+3}=\frac{4x+\frac{5}{x}+\frac{41}{3}}{x + 3}=\frac{\frac{12x^{2}+15 + 41x}{3}}{x + 3}=\frac{12x^{2}+41x + 15}{3(x + 3)}$.

Step2: Factor the numerator

Factor $12x^{2}+41x + 15$. We need to find two numbers that multiply to $12\times15 = 180$ and add up to $41$. The numbers are $5$ and $36$. So $12x^{2}+41x + 15=12x^{2}+36x+5x + 15=12x(x + 3)+5(x + 3)=(12x + 5)(x + 3)$.

Step3: Simplify the limit expression

$\frac{(12x + 5)(x + 3)}{3(x + 3)}=\frac{12x+5}{3}$ for $x\neq - 3$.

Step4: Evaluate the limit

$\lim_{x\rightarrow - 3}\frac{12x+5}{3}=\frac{12\times(-3)+5}{3}=\frac{-36 + 5}{3}=\frac{-31}{3}$.

The value of the limit is $\frac{31}{9}$.

The equation of a tangent - line is given by $y - y_{0}=m(x - x_{0})$, where $(x_{0},y_{0})=(-3,-\frac{41}{3})$ and $m = \frac{31}{9}$.

$y+\frac{41}{3}=\frac{31}{9}(x + 3)$

$y+\frac{41}{3}=\frac{31}{9}x+\frac{31}{3}$

$y=\frac{31}{9}x+\frac{31}{3}-\frac{41}{3}=\frac{31}{9}x-\frac{10}{3}$

Answer:

The value of the limit is $\frac{31}{9}$