Given $\displaystyle f(x)=\frac{2}{x-1}$ , use the four step process to find a slope-predicator function m(x). Then write an equation for the line tangent to the curve at the point $\displaystyle x = 0$.
I am not sure what you are asking.
If you want the tangent at $\displaystyle x=0$.
Use the slope-intercept form.
$\displaystyle y-y_0=m(x-x_0)$
Where,
$\displaystyle m=f'(0)$
And,
$\displaystyle x_0=0$ but then $\displaystyle y_0=\frac{2}{x_0-1}=-2$.
Also,
$\displaystyle f'(x)=-2(x-1)^{-2}$
At, $\displaystyle x=0$ we have,
$\displaystyle f'(0)=-2(0-1)^{-2}=-2$
Thus,
$\displaystyle y+2=-2(x-0)$
Thus,
$\displaystyle y=-2x-2$
Is equation of tangent line.
My text explains it as follows:
1. Write the definition of the derivative
2. Substitute the expresions f(x+h) and f(x) as determined by the paticular function of f.
3. Simplify the result by algebraic methods until it is possible to....
4. Apply appropriate limit laws to finally evaluate the limit.
Hello, FLTR!
Someone is inventing a new language ?!
Given $\displaystyle f(x) \:=\:\frac{2}{x-1}$, use the four-step process
to find a slope-predicator function m(x). . Why can't they say "derivative"?
Then write an equation for the line tangent to the curve at the point $\displaystyle x = 0$.
I assume your difficulty is with the algebra in the four-step process.
We have: .$\displaystyle f(x+h) - f(x)\;=\;\frac{2}{x+h-1} - \frac{2}{x-1}$
Get a common denominator:
. . $\displaystyle = \;\frac{2}{x+h-1}\cdot\frac{x-1}{x-1} \,- \,\frac{2}{x-1}\cdot\frac{x+h-1}{x+h-1} \;=\;\frac{2(x-1) - 2(x+h-1)}{(x-1)(x+h-1)}$
. . $\displaystyle = \;\frac{2x - 2 - 2x - 2h + 2}{(x-1)(x+h-1)} \;=\;\frac{-2h}{(x-1)(x+h-1)} $
Divide by $\displaystyle h\!:\;\;\frac{f+h) - f(x)}{h}\;=\;\frac{-2\not{h}}{\not{h}(x-1)(x+h-1)} \;= \;\frac{-2}{(x-1)(x+h-1)} $
Take limits: .$\displaystyle f'(x) \;= \;\lim_{h\to0}\frac{f(x+h) - f(x)}{h} \;=\;\lim_{h\to0}\frac{-2}{(x-1)(x+h-1)}$
Therefore: .$\displaystyle \boxed{f'(x)\;=\;\frac{-2}{(x-1)^2}}$
I assume you can finish the problem now . . .