What is the equation of the line normal to $f(x)= x^3/e^x$ at $x=2$ ?

Question

1447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-08T13:59:02

$y = 8/e^2 -e^2/4(x-2)$

The equation of the line normal to the graph of the function:

$y=f(x)$

at ath point $(bar x, f(barx))$ is given by:

$y=f(bar x)-1/(f'(barx)) (x-barx)$

In our case:

$f(x) = x^3/e^x=x^3e^(-x)$

$f(2) = 8/e^2$

$f'(x) = 3x^2e^-x -x^3e^-x$

$f'(2) = 12/e^2-8/e^2=4/e^2$

So the normal line is:

$y = 8/e^2 -e^2/4(x-2)$

What is the equation of the line normal to f(x)= x^3/e^x f(x)= x^3/e^x at x=2x=2?