# What is the equation of the tangent line of #f(x)=(2x+1)/(x+2) # at #x=1#?

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To find the equation of the tangent line of f(x) = (2x+1)/(x+2) at x=1, we need to find the slope of the tangent line and a point on the line.

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

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

Simplifying this expression, we get:

f'(x) = 3 / (x+2)^2

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

f'(1) = 3 / (1+2)^2 = 3/9 = 1/3

So, the slope of the tangent line at x=1 is 1/3.

To find a point on the line, we substitute x=1 into f(x):

f(1) = (2(1)+1)/(1+2) = 3/3 = 1

Therefore, the point on the line is (1, 1).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values we found, we get:

y - 1 = (1/3)(x - 1)

Simplifying this equation, we have:

y = (1/3)x + 2/3

Therefore, the equation of the tangent line of f(x) = (2x+1)/(x+2) at x=1 is y = (1/3)x + 2/3.

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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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