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

2y=

OR y=

f(x)= #sqrt(e^x-x)

The slope of the tangent to f(x) at x=1 is the first derivative at that point.

utilizing the line equation in point slope form

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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)

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