# How do you find the points where the graph of the function #y=2x(8-x)^.5# has horizontal tangents?

You compute the first derivative, set that equal to zero, solve for the x value(s), and then use the function to give you the corresponding y value(s).

Compute the first derivative:

Set that equal to zero:

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The curve

For the horizontal tgts., the slope of tgt., or, what is the same as to

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To find the points where the graph of the function y=2x(8-x)^.5 has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of the function y=2x(8-x)^.5 using the product rule and chain rule.

The derivative is given by: dy/dx = 2(8-x)^.5 + 2x(0.5)(8-x)^(-0.5)(-1)

Setting the derivative equal to zero, we have: 2(8-x)^.5 + 2x(0.5)(8-x)^(-0.5)(-1) = 0

Simplifying the equation, we get: (8-x)^.5 - x(8-x)^(-0.5) = 0

To solve this equation, we can square both sides: (8-x) - x^2(8-x)^(-1) = 0

Expanding and rearranging, we have: 8 - x - 8x^2 + x^3 = 0

This equation can be factored as: (x-4)(x^2 - 4x + 2) = 0

Setting each factor equal to zero, we get two possible values for x: x = 4 and x = 2 ± √2

Therefore, the points where the graph of the function y=2x(8-x)^.5 has horizontal tangents are (4, 0), (2 + √2, 0), and (2 - √2, 0).

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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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