How to find a equation that is tangent to #f(x)=1/sqrtx# and parallel to the line #-(1/2)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_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]}

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