What is the equation of the line normal to #f(x)=1/abs(x-6) # at #x=1#?
1 Answer
Jan 2, 2016
Or in general form:
#25x+y-126/5 = 0#
Explanation:
Unpack the definition of
#f(x) = { (1/(x-6), if x > 6), (-1/(x-6), if x < 6) :}#
So:
#f'(x) = { (-1/(x-6)^2, if x > 6), (1/(x-6)^2, if x < 6) :}#
In particular:
#f(1) = 1/5#
#f'(1) = 1/25#
So the slope of the tangent at
The equation of the normal can be written in point slope form as:
#y - 1/5 = -25 (x - 1)#
Hence:
#y = -25x+25+1/5#
That is:
#y = -25x + 126/5#
in slope intercept form.
Or in general form:
#25x+y-126/5 = 0#