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

1 Answer
May 31, 2016

78y=-x+4680

Explanation:

First find the slope of the tangent at the given point by differentiating and then substituting in x=4

f'(x)=8(x-2)^3 +3(x-1)^2 +2(x-3)

At x=4 this is m =8*2^3+3*2^2+2*1 = 64+12+2 = 78

The slope of the normal line is -1/m = -1/78

The equation of the normal is then y = -1/78x+c

To calculate c, find the value of the original function at the given point and then substitute into the equation of the normal.

f(4) = 2*2^4+3^3 +1^2 = 32+27+1 = 60

:.60=-1/78*4+c

c=60+4/78 = 4680/78

y=-1/78x+4680/78

:.78y=-x+4680