Question

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

Answer 1

`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.