What is the equation of the line normal to $f (x) = 2 x^{3} - x^{2} - x$ $f(x)=2x^3 - x^2-x$ at $x = 4$ $x=4$ ?

Question

1325 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-21T19:46:55

$R (x) = \frac{4 - x}{87} + 108$ $R(x) = {4-x}/87 + 108$

$f (4) = 2 \cdot 64 - 16 - 4 = 128 - 20 = 108$ $f(4) = 2 * 64 - 16 - 4 = 128 - 20 = 108$

$f'(x) = 6x ^2 - 2x - 1$

$f'(4) = 6*16-8-1=96-9=87$

$R(x) = -1/87 x + b$ is the desired perp.

$R(4) = 108 Leftrightarrow -4/87 + b = 108$

$b = 108 + 4/87$

What is the equation of the line normal to f(x)=2x^3 - x^2-x f(x)=2x3−x2−xf(x)=2x^3 - x^2-x at x=4x=4x=4?