# How do you find the equation of the line tangent to #f(x)= 1/x#, at (1/2,2)?

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graph{(1/x-y)(4x+y-4)=0 [-3.973, 4.8, -1.268, 3.117]}

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To find the equation of the line tangent to the function f(x) = 1/x at the point (1/2, 2), we can use the concept of differentiation.

First, we need to find the derivative of the function f(x). The derivative of 1/x can be found using the power rule for differentiation, which states that the derivative of x^n is n*x^(n-1). Applying this rule, the derivative of 1/x is -1/x^2.

Next, we substitute the x-coordinate of the given point (1/2, 2) into the derivative to find the slope of the tangent line. Plugging in x = 1/2 into the derivative -1/x^2, we get -1/(1/2)^2 = -1/(1/4) = -4.

Now that we have the slope of the tangent line, we can use the point-slope form of a linear equation to find the equation of the line. The point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. Substituting the values x1 = 1/2, y1 = 2, and m = -4, we get y - 2 = -4(x - 1/2).

Simplifying the equation, we have y - 2 = -4x + 2. Rearranging the terms, we get the equation of the line tangent to f(x) = 1/x at (1/2, 2) as y = -4x + 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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