# For what values of x does the graph of #f(x)=x^3+3x^2+x+3# have a horizontal tangent?

First, find the derivative.

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So, the tangents at these points (-0.1835, 2.911) and (-1.815, 5.089) are

horizontal.

The equations to these tangents are

y = 2.911 and y = 5.089.

graph{(y-2.911)(y-5.089)(y-3-x-3x^2-x^3)=0 [-20, 20, -10, 10]}

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To find the values of x where the graph of f(x) has a horizontal tangent, we need to find the points where the derivative of f(x) is equal to zero.

The derivative of f(x) is given by f'(x) = 3x^2 + 6x + 1.

Setting f'(x) equal to zero and solving for x, we get:

3x^2 + 6x + 1 = 0

Using the quadratic formula, we find the solutions for x:

x = (-6 ± √(6^2 - 4*3*1))/(2*3)

Simplifying further:

x = (-6 ± √(36 - 12))/(6)

x = (-6 ± √24)/(6)

x = (-6 ± 2√6)/(6)

Simplifying the expression:

x = -1 ± √6/3

Therefore, the values of x where the graph of f(x) has a horizontal tangent are x = -1 + √6/3 and x = -1 - √6/3.

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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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