How to find a equation that is tangent to f(x)=1/sqrtxf(x)=1√x and parallel to the line -(1/2)x+3−(12)x+3?

1 Answer
Shwetank Mauria
Aug 21, 2017

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_0x=x0 on the curve defined by function f(x)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]}

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