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 - 431))/(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.
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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