# What are the inflection points for #x^3 + 5x^2 + 4x - 3#?

When a function shifts from concave to convex (or vice versa), or when its concavity changes, it is referred to as an inflection point.

As a result, the second derivative

bringing the second derivative to zero (since the slope does not change at the inflection point; rather, it is the precise location where concavity changes, making it a derivative of zero):

diagram{x^3+5x^2+4x-3 [-10, 10, -5, 5]}

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To find the inflection points of the function ( f(x) = x^3 + 5x^2 + 4x - 3 ), we need to first find the second derivative, then find where it equals zero, and finally determine the corresponding ( x ) values.

The first derivative of ( f(x) ) is: [ f'(x) = 3x^2 + 10x + 4 ]

The second derivative is: [ f''(x) = 6x + 10 ]

Setting the second derivative equal to zero and solving for ( x ) gives: [ 6x + 10 = 0 ] [ x = -\frac{10}{6} = -\frac{5}{3} ]

So, the inflection point is at ( x = -\frac{5}{3} ).

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To find the inflection points of the function (f(x) = x^3 + 5x^2 + 4x - 3), we first need to find its second derivative. Then, we set the second derivative equal to zero and solve for the values of (x).

First derivative of (f(x)): [f'(x) = 3x^2 + 10x + 4]

Second derivative of (f(x)): [f''(x) = 6x + 10]

Setting (f''(x) = 0): [6x + 10 = 0] [6x = -10] [x = -\frac{10}{6}] [x = -\frac{5}{3}]

So, the inflection point of the function is (x = -\frac{5}{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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