# What is the equation of the line normal to  f(x)=xe^(sqrtx) at  x=1?

Jan 6, 2017

$y = - \frac{2}{3} {e}^{- 1} x + \frac{2}{3} {e}^{- 1} + e = - 0.245 x + 2.964$,, nearly.

See the normal-inclusive graph.

#### Explanation:

$y = x {e}^{\sqrt{x}}$ = e at x = 1.

So, the foot of the normal is P(1, e)..

Both $x \mathmr{and} y \ge 0 \mathmr{and} y \ge 0$.

$y ' = {e}^{\sqrt{x}} + \frac{1}{2} \sqrt{x} {e}^{\sqrt{x}}$=1.5e, at x = 1.

So, the slope of the normal is the negative reciprocal $- \frac{2}{3} {e}^{- 1}$.

And so, the equation to the normal at P(1, e) is

y-e=-3/3e&(-1)(x -1)#.

graph{(y-xe^(sqrtx))(y+0.245x-2.964)=0 [-10, 10, -5, 5]}