What is the equation of the line that is normal to #f(x)=4x^3-3x^2+5x-2# at # x=1 #?

1 Answer
Jan 26, 2016

#11y = -x +44#

Explanation:

The slope of the normal at a given point is the negative inverse of the slope of the function at that point. i.e. if the slope of the function is #m# then the slope of the normal is #-1/m#

To find the slope, differentiate the function.

#f'(x) = 12x^2 - 6x +5#

At #x=1# this equals # 12 -6 +5 = 11#

Therefore the slope of the normal is #-1/11#

The calculate the constant #c# in the standard lien equation #y=mx+c#, substitute #x=1# back into the original function #f(x)#
#=4*1^3 -3*1^2 +5*1 - 2 = 4#

Therefore the normal is #y = (-1/11)x +4#
#11y = -x +44#