use the definition of the derivative to find the derivative of the function $f(x)=\frac{3x + 5}{x - 1}$…

use the definition of the derivative to find the derivative of the function $f(x)=\frac{3x + 5}{x - 1}$. then find all $x$-values (if any) where the tangent line is horizontal. if the tangent line is horizontal for all $x$, write $mathbb{r}$ for your answer. if the tangent line is never horizontal, write none for your answer.

use the definition of the derivative to find the derivative of the function $f(x)=\frac{3x + 5}{x - 1}$. then find all $x$-values (if any) where the tangent line is horizontal. if the tangent line is horizontal for all $x$, write $mathbb{r}$ for your answer. if the tangent line is never horizontal, write none for your answer.

Answer

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\frac{3x + 5}{x - 1}$, then $f(x+h)=\frac{3(x + h)+5}{(x + h)-1}=\frac{3x+3h + 5}{x+h - 1}$.

Step2: Calculate $f(x + h)-f(x)$

[ \begin{align*} f(x + h)-f(x)&=\frac{3x+3h + 5}{x+h - 1}-\frac{3x + 5}{x - 1}\ &=\frac{(3x+3h + 5)(x - 1)-(3x + 5)(x+h - 1)}{(x+h - 1)(x - 1)}\ &=\frac{(3x^{2}-3x+3hx-3h + 5x-5)-(3x^{2}+3hx-3x+5x+5h - 5)}{(x+h - 1)(x - 1)}\ &=\frac{3x^{2}-3x+3hx-3h + 5x-5 - 3x^{2}-3hx + 3x-5x-5h + 5}{(x+h - 1)(x - 1)}\ &=\frac{-8h}{(x+h - 1)(x - 1)} \end{align*} ]

Step3: Calculate $\frac{f(x + h)-f(x)}{h}$

$\frac{f(x + h)-f(x)}{h}=\frac{\frac{-8h}{(x+h - 1)(x - 1)}}{h}=\frac{-8}{(x+h - 1)(x - 1)}$

Step4: Find the derivative

$f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h\rightarrow0}\frac{-8}{(x+h - 1)(x - 1)}=\frac{-8}{(x - 1)^{2}}$

Step5: Find when the tangent is horizontal

A horizontal tangent line occurs when $f^\prime(x)=0$. Set $\frac{-8}{(x - 1)^{2}}=0$. Since the numerator $- 8\neq0$ for all real - valued $x$, the equation $\frac{-8}{(x - 1)^{2}}=0$ has no solutions.

Answer:

None