What is the equation of the normal line of $f (x) = {(x - 3 e^{x})}^{\frac{1}{2}} - x^{2}$ $f(x)=(x-3e^x)^(1/2)-x^2$ at $x = - 3$ $x = -3$ ?

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What is the equation of the normal line of $f (x) = {(x - 3 e^{x})}^{\frac{1}{2}} - x^{2}$ $f(x)=(x-3e^x)^(1/2)-x^2$ at $x = - 3$ $x = -3$ ?

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Answer 1 · 2016-11-26T23:08:44

There is no normal

Explanation:

A normal can't be found as $f (x)$ $f(x)$ 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 $y = \sqrt{x}$ $y=sqrtx$ is $x > 0$ $x>0$ , but the range of $y = x - 3 e^{x}$ $y=x-3e^x$ is $y \leq - 2.099$ $y <= -2.099$ (3 d.p) - it is always negative.

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What is the equation of the normal line of $f (x) = {(x - 3 e^{x})}^{\frac{1}{2}} - x^{2}$ $f(x)=(x-3e^x)^(1/2)-x^2$ at $x = - 3$ $x = -3$ ?

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What is the equation of the normal line of f(x)=(x-3e^x)^(1/2)-x^2f(x)=(x−3ex)12−x2f(x)=(x-3e^x)^(1/2)-x^2 at x = -3x=−3x = -3?

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What is the equation of the normal line of $f (x) = {(x - 3 e^{x})}^{\frac{1}{2}} - x^{2}$ $f(x)=(x-3e^x)^(1/2)-x^2$ at $x = - 3$ $x = -3$ ?