What are the points of inflection, if any, of #f(x) =12x^5+15x^4–240x^3+6#?

To find the points of inflection of $f(x) = 12x^5 + 15x^4 - 240x^3 + 6$, we first need to find its second derivative, $f''(x)$. Then, we find the values of $x$ where $f''(x) = 0$ or $f''(x)$ does not exist. These values will be the potential points of inflection. After that, we check the concavity of the function around these points to confirm if they are indeed points of inflection.

First derivative of $f(x)$: $f'(x) = 60x^4 + 60x^3 - 720x^2$

Second derivative of $f(x)$: $f''(x) = 240x^3 + 180x^2 - 1440x$

Now, set $f''(x) = 0$: $240x^3 + 180x^2 - 1440x = 0$ $60x(x^2 + 3x - 24) = 0$ $60x(x + 6)(x - 4) = 0$

This gives us potential points of inflection at $x = 0, x = -6,$ and $x = 4$.

Now, we check the concavity of $f(x)$ around these points:

1. For $x = 0$: Substitute $x = 0$ into $f''(x)$ to find the concavity: $f''(0) = 0 + 0 - 0 = 0$ So, $x = 0$ is a point of inflection.

2. For $x = -6$: Substitute $x = -6$ into $f''(x)$ to find the concavity: $f''(-6) = -1440(-6) < 0$ So, $x = -6$ is a point of inflection.

3. For $x = 4$: Substitute $x = 4$ into $f''(x)$ to find the concavity: $f''(4) = 240(4)^3 + 180(4)^2 - 1440(4) > 0$ So, $x = 4$ is a point of inflection.

Therefore, the points of inflection for the function $f(x) = 12x^5 + 15x^4 - 240x^3 + 6$ are $x = 0, x = -6,$ and $x = 4$.