Question

What is the equation of the normal line of `f(x)= x^e*e^x` at `x=3`?

Answer 1

`y-3^ee^3=-(3^(1-e))/(e^3(e+3))(x-3)`

Explanation:

First, find the point the normal line will intercept.

`f(3)=3^ee^3`

The normal line will pass through the point `(3,3^ee^3)`.

To find the slope of the normal line, we should first find the derivative of the function. This can be found through the product rule:

`f'(x)=e^xd/dx[x^e]+x^ed/dx[e^x]`

Note that `d/dx[e^x]=e^x` and that, through the power rule, `d/dx[x^e]=ex^(e-1)`.

Thus,

`f'(x)=e^x(ex^(e-1))+x^ee^x`

`f'(x)=e^x(ex^(e-1)+x^e)`

The slope of the tangent line at `x=3` is

`f'(3)=e^3(e(3^(e-1))+3^e)`

`f'(3)=e^3 3^(e-1)(e+3)`

However, we want to find the normal line of the function. The slopes of the tangent line and normal line are perependicular, to they are opposite reciprocals. The opposite reciprocal of `f'(3)` is

`-(e^3 3^(e-1)(e+3))^-1=-(3^(1-e))/(e^3(e+3))`

The slope of `-(3^(1-e))/(e^3(e+3))` and point of `(3,3^ee^3)` can be related as an equation in point slope form:

`y-3^ee^3=-(3^(1-e))/(e^3(e+3))(x-3)`