# What are the points of inflection, if any, of #f(x) = x^6 + 3x^5 - (15/2)x^4 - 40x^3 - 60x^2 + 8x + 5 #?

Points of inflection occur at

The graph is shrunk vertically and is not drawn to scale: graph{x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5 [-4, 4, -200, 200]}

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To find the points of inflection of ( f(x) = x^6 + 3x^5 - \frac{15}{2}x^4 - 40x^3 - 60x^2 + 8x + 5 ), we first need to find the second derivative, ( f''(x) ). Then, we solve ( f''(x) = 0 ) to find the potential points of inflection. Finally, we analyze the sign changes in ( f''(x) ) around those potential points to determine if they are points of inflection.

First derivative: ( f'(x) = 6x^5 + 15x^4 - 15x^3 - 120x^2 - 120x + 8 )

Second derivative: ( f''(x) = 30x^4 + 60x^3 - 45x^2 - 240x - 120 )

To find the potential points of inflection, solve ( f''(x) = 0 ) for ( x ): ( 30x^4 + 60x^3 - 45x^2 - 240x - 120 = 0 )

Once you find the solutions for ( x ), you can analyze the sign changes of ( f''(x) ) around those points to determine the points of inflection. If the sign changes from positive to negative or vice versa, the point is a point of inflection.

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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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