Question

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

Answer 1

`y=-2/3e^(-1)x+2/3e^(-1)+e=-0.245x+2.964`,, nearly.

See the normal-inclusive graph.

Explanation:

`y=xe^sqrtx` = e at x = 1.

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

Both `x and y >=0 and y >=0`.

`y'=e^sqrtx+1/2sqrtxe^sqrtx`=1.5e, at x = 1.

So, the slope of the normal is the negative reciprocal `-2/3e^(-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]}