# How do you find the equation of the tangent line to the curve at the given point. #y=5x−2sqrtx#, #(1,3)#?.

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To find the equation of the tangent line to the curve at the given point (1,3), we need to find the derivative of the function y=5x−2√x and evaluate it at x=1. The derivative of the function is dy/dx = 5 - √x. Evaluating this at x=1, we get dy/dx = 5 - √1 = 5 - 1 = 4.

The slope of the tangent line is equal to the derivative evaluated at the given point, so the slope is 4.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values (1,3) and m=4 into the equation.

Therefore, the equation of the tangent line to the curve y=5x−2√x at the point (1,3) is y - 3 = 4(x - 1).

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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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