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 derivative $f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}$. Here $f(x)=4x+\frac{5}{x}$, $a = - 3$, and $f(-3)=4\times(-3)+\frac{5}{-3}=-12-\frac{5}{3}=-\frac{36 + 5}{3}=-\frac{41}{3}\approx - 13.666\cdots$. [ \begin{align*} f'(-3)&=\lim_{x\rightarrow - 3}\frac{4x+\frac{5}{x}-(-\frac{41}{3})}{x + 3}\ &=\lim_{x\rightarrow - 3}\frac{4x+\frac{5}{x}+\frac{41}{3}}{x + 3}\ &=\lim_{x\rightarrow - 3}\frac{12x^{2}+15 + 41x}{3x(x + 3)}\ &=\lim_{x\rightarrow - 3}\frac{12x^{2}+41x + 15}{3x(x + 3)} \end{align*} ] Factor the numerator: $12x^{2}+41x + 15=(3x + 1)(4x+15)$. [ \begin{align*} f'(-3)&=\lim_{x\rightarrow - 3}\frac{(3x + 1)(4x + 15)}{3x(x + 3)}\ &=\lim_{x\rightarrow - 3}\frac{12x^{2}+45x+4x + 15}{3x^{2}+9x}\ &=\lim_{x\rightarrow - 3}\frac{12x^{2}+49x + 15}{3x^{2}+9x}\ &=\lim_{x\rightarrow - 3}\frac{(x + 3)(12x+5)}{3x(x + 3)}\ &=\lim_{x\rightarrow - 3}\frac{12x+5}{3x} \end{align*} ]

Step2: Evaluate the limit

Substitute $x=-3$ into $\frac{12x + 5}{3x}$: [ \frac{12\times(-3)+5}{3\times(-3)}=\frac{-36 + 5}{-9}=\frac{-31}{-9}=\frac{31}{9}\approx3.444 ] The equation of the tangent - line is given by $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-3,-\frac{41}{3})$ and $m = f'(-3)=\frac{31}{9}$. [ \begin{align*} 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}\ y&=\frac{31}{9}x-\frac{10}{3} \end{align*} ]

Answer:

The value of $f'(-3)$ is $\frac{31}{9}$ and the equation of the tangent line is $y=\frac{31}{9}x-\frac{10}{3}$