Question

if $f(x)=4x+\frac{5}{x}$, find $f(-3)$, using the definition of derivative. $f(-3)$ is the limit as $x\to$ of the expression. the value of this limit is $\frac{31}{9}$. use this to find the equation of the tangent line to the graph of $y = 4x+\frac{5}{x}$ at the point $(-3,-13.6666666666667)$. the equation of this tangent line can be written in the form $y=\frac{31}{9}x-\frac{10}{3}$

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function $y = f(x)$ at $x = a$ is $f^{\prime}(a)=\lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}$. Here, $f(x)=4x+\frac{5}{x}$, $a=-3$, and $f(-3)=4\times(-3)+\frac{5}{-3}=-12-\frac{5}{3}=-\frac{36 + 5}{3}=-\frac{41}{3}\approx - 13.6666666667$. 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 the quadratic $12x^{2}+41x + 15$. 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

The expression $\frac{12x^{2}+41x + 15}{3(x + 3)}$ becomes $\frac{(12x + 5)(x + 3)}{3(x + 3)}$. Canceling out the common factor $(x + 3)$ (for $x
eq - 3$), we get $\frac{12x+5}{3}$.

Step4: Evaluate the limit

Now, find $\lim_{x
ightarrow - 3}\frac{12x+5}{3}$. Substitute $x=-3$ into $\frac{12x+5}{3}$, we have $\frac{12\times(-3)+5}{3}=\frac{-36 + 5}{3}=\frac{-31}{3}$.

The equation of a tangent - line to the curve $y = f(x)$ at the point $(x_0,y_0)$ is given by $y - y_0=f^{\prime}(x_0)(x - x_0)$. Here $x_0=-3$, $y_0=-\frac{41}{3}$, and $f^{\prime}(-3)=\frac{-31}{3}$. The equation of the tangent line is $y+\frac{41}{3}=\frac{-31}{3}(x + 3)$. Multiply through by 3 to get $3y+41=-31x-93$, or $y=-\frac{31}{3}x-\frac{93 + 41}{3}=-\frac{31}{3}x-\frac{134}{3}\approx-\frac{31}{3}x-44.6666666667$.

Answer:

$f^{\prime}(-3)=-\frac{31}{3}$; The equation of the tangent line is $y =-\frac{31}{3}x-\frac{134}{3}$