We have: #f(x) = x^(3) - 3 x^(2)#
First, let's find the #y#-coordinate corresponding to the given value of #x#:
#Rightarrow f(4) = (4)^(3) - 3 cdot (4)^(2)#
#Rightarrow f(4) = 64 - 48#
#therefore f(4) = 16#
So we must find the equation of the normal line at the point #(4, 16)#.
Then, let's differentiate #f(x)#:
#Rightarrow f'(x) = 3 x^(2) - 6 x#
For #x = 4#, the slope of the line is:
#Rightarrow f'(4) = 3 cdot (4)^(2) - 6 cdot (4)#
#Rightarrow f'(4) = 48 - 24#
#therefore f'(4) = 24#
The gradient of the normal line is the reciprocal of this value:
#Rightarrow "Gradient of normal line" = (f'(4))^(- 1)#
#therefore "Gradient of normal line" = frac(1)(24)#
Now, let's express the equation of the normal line in point-slope form:
#Rightarrow y - y_(1) = m (x - x_(1))#
#Rightarrow y - 16 = frac(1)(24) (x - 4)#
#Rightarrow y - 4 = frac(1)(24) x - frac(1)(6)#
#therefore y = frac(1)(24) x - frac(23)(6)#
Therefore, the equation of the normal line is #y = frac(1)(24) x - frac(23)(6)#.
