What is the equation of the normal line of $f (x) = e^{- x} + x^{3}$ $f(x)=e^-x+x^3$ at $x = - 5$ $x=-5$ ?

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Answer 1 · 2016-10-31T11:25:51

$y = - \frac{x}{e^{5} - 75} + e^{5} - 125 - \frac{5}{e^{5} - 75}$ $y = -x/(e^5-75) +e^5-125-5/(e^5-75)$

$f (x) = e^{- x} + x^{3}$ $f(x)=e^-x+x^3$

When $x = - 5 \Rightarrow f (- 5) = e^{5} - 125$ $x=-5 => f(-5)=e^5-125$

Differentiating wrt $x$ $x$ ;
$f (x) = - e^{- x} + 3 x^{2}$ $f(x)=-e^-x+3x^2$
so When $x=-5 => f'(-5)=-e^5+75 =75-e^5$

This is the slope of the tangent when $x=-5$ , as the normal is perpendicular to the tangent, their product is $-1$

So, the slope of the normal is $-1/(75-e^5)=1/(e^5-75)$

Using $y-y_1=m(x-x_1)$ , the normal equation is;

$y-(e^5-125) = 1/(e^5-75)(x-(-5))$
$y-e^5+125 = 1/(e^5-75)(x+5)$
$y-e^5+125 = x/(e^5-75)+5/(e^5-75)$
$y = x/(e^5-75) +e^5-125+5/(e^5-75)$

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What is the equation of the normal line of f(x)=e^-x+x^3f(x)=e−x+x3f(x)=e^-x+x^3 at x=-5x=−5x=-5?