What is the equation of the normal line of #f(x)=sqrt(e^x-x+1)/e^(2x)# at #x = 0#?

Question

Charles Arocha · Accepted Answer

The slope of the tangent line is the first derivative evaluated at the the given x coordinate. The slope of the normal line is the negative of the reciprocal of the slope of the tangent line.

Compute the first derivative: #f'(x) = {e^(-2x)(e^x - 1)}/{2sqrt(e^x - x + 1)} - 2e^(2x)sqrt(e^x - x + 1)# Evaluate at x = 0: #f'(0) = {e^(-2(0))(e^0 - 1)}/{2sqrt(e^0 - 0 + 1)} - 2e^(2(0))sqrt(e^0 - 0 + 1)# #f'(0) = {(1)(1 - 1)}/{2sqrt(1 + 1)} - sqrt(1 + 1)# #f'(0) = -sqrt(2)# The slope of the normal line is: #-1/-sqrt(2) = sqrt(2)/2# Given #x = 0# we observe that we are computing the y intercept when we evaluate f(0): #f(0) = sqrt(2)# Using the slope-intercept form of the equation of a line: #y = (sqrt(2)/2)x + sqrt(2)#

Elijah Abram · Answer

undefined To find the equation of the normal line of f(x) at x = 0, we need to determine the slope of the tangent line at x = 0 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) using the quotient rule:

f'(x) = [ (e^x - x + 1) * (e^(2x))' - (e^(2x)) * (e^x - x + 1)' ] / (e^(2x))^2

Simplifying this expression, we get:

f'(x) = [ (e^x - x + 1) * (2e^(2x)) - (e^(2x)) * (e^x - 1) ] / (e^(2x))^2

Now, we substitute x = 0 into f'(x) to find the slope of the tangent line at x = 0:

f'(0) = [ (e^0 - 0 + 1) * (2e^(2*0)) - (e^(2*0)) * (e^0 - 1) ] / (e^(2*0))^2

Simplifying further, we have:

f'(0) = [ (1) * (2) - (1) * (0) ] / (1)^2

f'(0) = 2

The slope of the tangent line at x = 0 is 2.

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

m_normal = -1/2

Now, we have the slope of the normal line. To find the equation of the normal line, we use the point-slope form of a line and substitute the point (0, f(0)) into the equation:

y - f(0) = m_normal * (x - 0)

Simplifying, we get:

y - f(0) = -1/2 * x

Finally, we can substitute f(0) into the equation using the given function f(x):

y - sqrt(e^0 - 0 + 1)/e^(2*0) = -1/2 * x

Simplifying further, we have:

y - 1 = -1/2 * x

This is the equation of the normal line of f(x) at x = 0.