Question

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

Answer 1

`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`