Question

What is the equation of the line tangent to `f(x)= -1/(3-2x)` at `x=-3`?

Answer 1

81y + 2x - 3 = 0

Explanation:

The equation of the tangent is: y - b = m(x - a )

where m represents the gradient (slope) ,(a , b ) is a point on
the line.

Require to find m and (a , b ). the derivative f'(x) = m and

the point (a ,b ) by using x = - 3

rewrite f(x) `= - (3 - 2x )^-1`

differentiate using `color(blue)("chain rule")`

`f'(x) = (3 - 2x )^-2 .d/dx (3 - 2x )`

`= (3 - 2x )^-2 .(-2 ) = - 2/(3 - 2x )^2`

x`= - 3 : f'( - 3 ) = - 2/(3 + 6 )^2 = -2/81`

and y `= 1/(3 + 6 ) = 1/9`

equation is then : y - `1/9 =( -2)/81 (x + 3 )`

( multiply thro' by 81 to eliminate fractions )

hence 81y - 9 = - 2 (x + 3 )

and 81y - 9 = - 2x - 6

`rArr 81y + 2x - 3 = 0`