What is the equation of the line normal to $f (x) = \frac{x^{2}}{e^{x}} - \frac{x}{e^{x^{2}}}$ $f(x)=x^2/e^x-x/e^(x^2)$ at $x = 0$ $x=0$ ?

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Answer 1 · 2015-12-29T07:04:41

$y = x$ $y = x$

$Graph of f (x)$ $color(blue)("Graph of "f(x))$
graph{(x^2)/(e^x)-x/(e^(x^2)}

$f (0) = 0$ $f(0) = 0$

Since the Normal passes through $(0, 0)$ $(0,0)$ , it has the form of $y = m x$ $y = mx$ .

Since the Normal is perpendicular to the tangent of $f$ $f$ at $x = 0$ $x = 0$ , $m*f'(0) = -1$ .

To find $f'(0)$ , we first find the general $f'(x)$ , and then substitute $x = 0$ .

$f'(x) = frac{d}{dx}( frac{x^2}{e^x} - frac{x}{ e^{x^2} } )$

$= frac{e^x(2x) - x^2(e^x)}{(e^x)^2} - frac{e^{x^2}(1) - x(2xe^{x^2})}{(e^{x^2})^2}$

$= frac{2x - x^2}{e^x} - frac{1 - 2x^2}{e^{x^2}}$

$f'(0) = -1$

Therefore, $m = frac{-1}{-1} = 1$

Equation of Normal: $y = x$

$color(blue)("Graph of Normal")$
graph{x}

What is the equation of the line normal to f(x)=x^2/e^x-x/e^(x^2) f(x)=x2ex−xex2 f(x)=x^2/e^x-x/e^(x^2) at x=0x=0 x=0?