What is the equation of the line normal to #f(x)= x^3-e^x # at #x=2#?

1 Answer

#y = (2-x)/(12 - e^2) -8 + e^2#

Explanation:

First, we have to Find the Required Coordinate on which the Normal Line will intersect with the graph.

So, We are given that the point has #x = 2#.

So, The required point is #(2, f(2)) = (2, 2^3 - e^2) = (2, 8 - e^2)#.............(i)

Now, We need to calculate the slope of the graph on that point.

So, The equation of slope for #f(x)# is:

#f'(x) = d/dxf(x) = d/dx(x^3 - e^x) = d/dx x^3 - d/dx e^x = 3x^2 - e^x#

Now. we will just substitute #x = 2# in #f'(x)# to get the slope.

So, The slope of #f(x)# at #x = 2# is:

#f'(2) = 3xx 2^2 - e^2 = 12 - e^2#.

Now, we know, Condition of Perpendicularity of two lines is #m_1m_2 = -1# where #m_1# and #m_2# are the slopes of the lines respectively.

So, The slope of our normal line is #-1/(12 - e^2)#.

Now, Let's find the equation.

We know the Slope-Intercept Form of a linear equation is:

#y = mx + c# [Where, #m# is the slope and #c# is the y-intercept.]

In this equation, let's substitute our slope.

So, The equation becomes,

#y = -1/(12 - e^2) xx x + c rArr y = -x/(12 - e^2) + c#.........................(ii)

And, This line must pass through our desired point #(2, 8 -e^2)#.

So, the equation must be satisfied with those values.

So, Now substituting the values of #x# and #y#, solve for #c#.

So, We get,

#8 - e^2 = -2/(12 - e^2) + c rArr c = 2/(12 - e^2) -8 + e^2#.

Now, We will put #c# in eq (ii) to get the answer.

So, We get,

#y = (2-x)/(12 - e^2) -8 + e^2# ........................(iii)

Hope this helps.