What is the equation of the normal line of f(x)=e^-x+x^3f(x)=ex+x3 at x=-5x=5?

1 Answer
Oct 31, 2016

y = -x/(e^5-75) +e^5-125-5/(e^5-75)y=xe575+e51255e575

Explanation:

f(x)=e^-x+x^3 f(x)=ex+x3

When x=-5 => f(-5)=e^5-125 x=5f(5)=e5125

Differentiating wrt xx;
f(x)=-e^-x+3x^2 f(x)=ex+3x2
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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