What is the equation of the line tangent to # f(x)=sqrt(e^x-x) # at # x=1#?

Question

What is the equation of the line tangent to #  f(x)=sqrt(e^x-x)  # at #  x=1#?

Samantha Adkison · Accepted Answer

2y=#sqrt(e-1)x +sqrt(e-1)#
OR y=#sqrt(e-1)x/2+sqrt(e-1)/2# f(x)= #sqrt(e^x-x) The slope of the tangent to f(x) at x=1 is the first derivative at that point. f'(x)= #(e^x-1)/(2sqrt(e^x-x))(using [chain rule](https://tutor.hix.ai) ) # f'(1)=#(e-1)/(2sqrt(e-1))# f'(1) =#(e-1)/2# slope =#(e-1)/2# and points (1,f(1) ) ; f(1)=#sqrt(e-1)# utilizing the line equation in point slope form y-y1=m(x-x1) ; m=slope=#(e-1)/2# ;x1=1 and y1=#sqrt(e-1)# y-#sqrt(e-1)#=#(e-1)/2#(x-1) 2y-2#sqrt(e-1)#=#sqrt(e-1)#x-#sqrt(e-1)# 2y=#sqrt(e-1)#x+#sqrt(e-1)# y=#sqrt(e-1)x/2+sqrt(e-1)/2#

Paisley Johnson · Answer

undefined To find the equation of the line tangent to the function f(x) = √(e^x - x) at x = 1, we need to determine the slope of the tangent line at that point and the coordinates of the point itself.

First, let's find the derivative of f(x) with respect to x.

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

Next, substitute x = 1 into f'(x) to find the slope of the tangent line at x = 1.

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

Now, evaluate f(1) to find the y-coordinate of the point of tangency.

f(1) = √(e^1 - 1)

Therefore, the equation of the line tangent to f(x) = √(e^x - x) at x = 1 is:

y - f(1) = f'(1) * (x - 1)