# How do you find the equation of the tangent line to the graph of #f(x)=sqrt(x-1)# at point #(5,2)#?

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To find the equation of the tangent line to the graph of f(x) = √(x-1) at point (5,2), we need to find the slope of the tangent line and the coordinates of the point.

First, we find the derivative of f(x) with respect to x, which is f'(x) = 1 / (2√(x-1)).

Next, we substitute x = 5 into f'(x) to find the slope of the tangent line at x = 5.

f'(5) = 1 / (2√(5-1)) = 1 / (2√4) = 1 / (2 * 2) = 1/4.

So, the slope of the tangent line at x = 5 is 1/4.

Now, we have the slope and the point (5,2). We can use the point-slope form of a linear equation to find the equation of the tangent line.

Using the point-slope form, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values.

y - 2 = (1/4)(x - 5).

Simplifying the equation, we get the equation of the tangent line to the graph of f(x) = √(x-1) at point (5,2) as y = (1/4)x + 9/4.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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