# How do you find the equation of the line tangent to the graph of #f(x) =10/(5x + 9#), when x =1/5?

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To find the equation of the line tangent to the graph of f(x) = 10/(5x + 9) at x = 1/5, we need to find the derivative of f(x) and evaluate it at x = 1/5. The derivative of f(x) is found using the quotient rule, which states that the derivative of a quotient of two functions is equal to (f'(x)g(x) - f(x)g'(x))/(g(x))^2. Applying the quotient rule to f(x) = 10/(5x + 9), we get f'(x) = -50/(5x + 9)^2. Evaluating f'(x) at x = 1/5, we have f'(1/5) = -50/(5(1/5) + 9)^2 = -50/(1 + 9)^2 = -50/100 = -1/2. Therefore, the slope of the tangent line is -1/2. Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute x1 = 1/5, y1 = f(1/5) = 10/(5(1/5) + 9) = 10/10 = 1 into the equation. Simplifying, we get y - 1 = (-1/2)(x - 1/5). This can be further simplified to y = -1/2x + 11/10, which is the equation of the line tangent to the graph of f(x) = 10/(5x + 9) at x = 1/5.

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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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