What is the equation of the line normal to # f(x)=e^(sqrtx-x)# at # x=4#?

Question

What is the equation of the line normal to #  f(x)=e^(sqrtx-x)# at #  x=4#?

Charles Anctil · Accepted Answer

I get #y = (4e^2)/3x-(16e^2)/3+1/e^2# #f(x) = e^(sqrtx-x)# #f(4) = e^-2 = 1/e^2# #f'(x) = e^(sqrtx-x)(1/(2sqrtx)-1)# #f'(4) =e^(sqrt4-4)(1/(2sqrt4)-1) = e^(2-4)(1/4-1) = e^(-2)(-3/4) = -3/(4e^2)# The normal line it perpendicular to the tangent, so the slope of the normal line is #-1/(f'(4)) = (4e^2)/3#. The line through #(4,1/e^2 )# with slope #m=(4e^2)/3# is #y-1/e^2 = (4e^2)/3(x-4)# or #y = (4e^2)/3x-(16e^2)/3+1/e^2#

Cameron Alexander · Answer

undefined To find the equation of the line normal to the function f(x) = e^(sqrt(x) - x) at x = 4, we need to determine the slope of the tangent line at x = 4 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we differentiate the function f(x) with respect to x.

f'(x) = (1/2) * e^(sqrt(x) - x) * (1/sqrt(x)) - e^(sqrt(x) - x)

Evaluating f'(4), we substitute x = 4 into the derivative expression:

f'(4) = (1/2) * e^(sqrt(4) - 4) * (1/sqrt(4)) - e^(sqrt(4) - 4)

Simplifying further:

f'(4) = (1/2) * e^(2 - 4) * (1/2) - e^(2 - 4)
      = (1/2) * e^(-2) * (1/2) - e^(-2)
      = (1/4) * e^(-2) - e^(-2)
      = (-3/4) * e^(-2)

The slope of the tangent line at x = 4 is -3/4 * e^(-2).

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

Slope of the normal line = -1 / (-3/4 * e^(-2))
                       = 4/3 * e^2

Now that we have the slope of the normal line, we can use the point-slope form of a line to find the equation. Since the normal line passes through the point (4, f(4)), we substitute x = 4 into the original function:

f(4) = e^(sqrt(4) - 4)
     = e^(2 - 4)
     = e^(-2)

Therefore, the point of intersection is (4, e^(-2)).

Using the point-slope form, we have:

y - e^(-2) = (4/3 * e^2) * (x - 4)

Simplifying further:

y = (4/3 * e^2) * (x - 4) + e^(-2)

This is the equation of the line normal to f(x) = e^(sqrt(x) - x) at x = 4.