Question

What is the equation of the normal line of `f(x)=(x-3e^x)^(1/2)-x^2` at `x = -3`?

Answer 1

There is no normal

Explanation:

A normal can't be found as `f(x)` doesn't exist as the square root in the formula is always negative and you can't have a negative square root (that is real at least). graph{y=x-3e^x [-40.73, 38.1, -26.78, 12.62]}

The domain of `y=sqrtx` is `x>0`, but the range of `y=x-3e^x` is `y <= -2.099` (3 d.p) - it is always negative.