# What are the points of inflection, if any, of #f(x)=x^4/(x^3+6 #?

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To find the points of inflection for ( f(x) = \frac{x^4}{x^3 + 6} ), we first need to find the second derivative and then solve for the points where the second derivative is zero or undefined.

The first derivative of ( f(x) ) is: [ f'(x) = \frac{d}{dx} \left( \frac{x^4}{x^3 + 6} \right) ] Using the quotient rule, we get: [ f'(x) = \frac{(4x^3)(x^3 + 6) - x^4(3x^2)}{(x^3 + 6)^2} ] Simplifying further: [ f'(x) = \frac{4x^6 + 24x^3 - 3x^6}{(x^3 + 6)^2} ] [ f'(x) = \frac{x^6 + 24x^3}{(x^3 + 6)^2} ]

Now, to find the second derivative, we differentiate ( f'(x) ): [ f''(x) = \frac{d}{dx} \left( \frac{x^6 + 24x^3}{(x^3 + 6)^2} \right) ] Using the quotient rule and simplifying: [ f''(x) = \frac{(6x^5 + 72x^2)(x^3 + 6)^2 - (x^6 + 24x^3)(2(x^3 + 6)(3x^2))}{(x^3 + 6)^4} ] [ f''(x) = \frac{6x^5(x^3 + 6)^2 + 72x^2(x^3 + 6)^2 - 6x^5(x^3 + 6)^2 - 288x^5(x^3 + 6)}{(x^3 + 6)^4} ] [ f''(x) = \frac{72x^2(x^3 + 6)^2 - 288x^5(x^3 + 6)}{(x^3 + 6)^4} ] [ f''(x) = \frac{72x^2(x^3 + 6)(x^3 + 6 - 4x^3)}{(x^3 + 6)^4} ] [ f''(x) = \frac{72x^2(x^3 + 6)(x^3 - 6)}{(x^3 + 6)^4} ]

The points of inflection occur where ( f''(x) = 0 ) or is undefined. Since the denominator ( (x^3 + 6)^4 ) is never zero, we only need to find where the numerator ( 72x^2(x^3 + 6)(x^3 - 6) ) is zero: [ 72x^2(x^3 + 6)(x^3 - 6) = 0 ]

The critical points are ( x = 0 ), ( x = \sqrt[3]{6} ), and ( x = -\sqrt[3]{6} ). However, ( x = 0 ) does not satisfy ( f''(x) = 0 ) as it makes the numerator non-zero. Therefore, the points of inflection are at ( x = \sqrt[3]{6} ) and ( x = -\sqrt[3]{6} ).

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