What is the equation of the normal line of #f(x)=e^x-x^3# at #x=0#?

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Answer 1 · 2015-12-20T15:11:24

In slope intercept form:

#y = -x+1#

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

Then #f'(x) = e^x-3x^2#

#f(0) = e^0 - 0 = 1#

So the graph of the function passes through #(0, 1)#

#f'(0) = e^0 - 0 = 1#

So the slope of the tangent at #(0, 1)# is #1#.

If the slope of the tangent is #m# then the slope of the normal is #-1/m#. So in our example the slope of the normal is #-1/1 = -1#.

The equation of a line of slope #m# passing through a point #(x_0, y_0)# can be written as:

#(y - y_0) = m(x - x_0)#

In our case we have #(x_0, y_0) = (0, 1)# and #m = -1#, so:

#(y - 1) = -1(x - 0) = -x#

That is #y = -x+1# in slope intercept form.

graph{(e^x-x^3-y)(x+y-1) = 0 [-10, 10, -5, 5]}