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

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Answer 1 · 2017-01-06T16:50:52

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

See the normal-inclusive graph.

#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]}