Question

How to find a equation that is tangent to `f(x)=1/sqrtx` and parallel to the line `-(1/2)x+3`?

Answer 1

Please see below.

Explanation:

The slope of the tangent at a point is the value of the first derivative at that point i.e. if we are seeking a tangent at `x=x_0` on the curve defined by function `f(x)`, the slope is `f'(x_0)` and hence equation of tangent is

`(y-f(x_0))=f'(x_0)(x-x_0)`

Here given function is `f(x)=1/sqrtx` and hence `f'(x)=-1/2xx1/(x^(3/2))=-1/(2sqrt(x^3))`

As tangent is parallel to line `y=-1/2x+3`, slope of tangent should be `-1/2` and hence `-1/(2sqrt(x^3))=-1/2` i.e. `sqrt(x^3)=1` or `x=1` and at `x=1`, `f(1)=1`.

Hence tangent is `y-1=-1/2(x-1)` or `2y-2=-x+1` i.e. `x+2y-3=0`

graph{(y-1/sqrtx)(2y+x-6)(x+2y-3)=0 [-1.275, 8.725, -1.15, 3.85]}