# How do you find the points where the graph of the function #y=(4x)/(x+1)# has horizontal tangents and what is the equation?

Find the derivative, set it to

Use the quotient rule:

A different example

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To find the points where the graph of the function y=(4x)/(x+1) 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=(4x)/(x+1) using the quotient rule. The derivative is given by:

dy/dx = [(4)(x+1) - (4x)(1)] / (x+1)^2

Next, we set the derivative equal to zero and solve for x:

[(4)(x+1) - (4x)(1)] / (x+1)^2 = 0

Simplifying the equation, we have:

4(x+1) - 4x = 0

Expanding and combining like terms:

4x + 4 - 4x = 0

4 = 0

Since the equation 4 = 0 is not true, there are no values of x that satisfy this equation. Therefore, the graph of the function y=(4x)/(x+1) does not have any points where it has horizontal tangents.

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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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