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

Question

1473 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T11:25:51

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

# f(x)=e^-x+x^3 #

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

Differentiating wrt #x#;
# 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) #

enter image source here