Consider the function $$ f(x)=12 x^{5}+60 x^{4}-240 x^{3}+3 $$. $$ f(x) $$ has inflection points at (reading from left to right) $$ x=D, E $$, and $$ F $$ where $$ D $$ is and $$ E $$ is and $$ F $$ is

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We begin with the function

$$
f(x)=12x^5+60x^4-240x^3+3.
$$

**Step 1. Find the Second Derivative**

The first derivative is

$$
f'(x) = \frac{d}{dx}(12x^5) + \frac{d}{dx}(60x^4) - \frac{d}{dx}(240x^3) + \frac{d}{dx}(3) = 60x^4 + 240x^3 - 720x^2.
$$

The second derivative is then

$$
f''(x)=\frac{d}{dx}(60x^4) + \frac{d}{dx}(240x^3) - \frac{d}{dx}(720x^2)=240x^3+720x^2-1440x.
$$

We can factor out a common factor:

$$
f''(x)=240x(x^2+3x-6).
$$

**Step 2. Solve for Inflection Points**

Inflection points occur where $$ f''(x)=0 $$ and the concavity changes. Set

$$
240x(x^2+3x-6)=0.
$$

This gives two factors:

1. $$ 240x=0 $$ ⟹ $$ x=0 $$.

2. $$ x^2+3x-6=0 $$.

Solve the quadratic equation using the quadratic formula:

$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{3^2-4(1)(-6)}}{2(1)}=\frac{-3\pm\sqrt{9+24}}{2}=\frac{-3\pm\sqrt{33}}{2}.
$$

Thus, the three $$ x $$-values where $$ f''(x)=0 $$ are

$$
x=\frac{-3-\sqrt{33}}{2}, \quad x=0, \quad x=\frac{-3+\sqrt{33}}{2}.
$$

**Step 3. Order the Inflection Points**

Reading from left to right (i.e. from the smallest $$ x $$ value to the largest):

- $$ D=\frac{-3-\sqrt{33}}{2} $$
- $$ E=0 $$
- $$ F=\frac{-3+\sqrt{33}}{2} $$
The inflection points of the function $$ f(x)=12x^5+60x^4-240x^3+3 $$ are at: - $$ D = \frac{-3 - \sqrt{33}}{2} $$ - $$ E = 0 $$ - $$ F = \frac{-3 + \sqrt{33}}{2} $$

Consider the function
.
has inflection points at (reading from left to right)
, and
where is
and is
and is

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Consider the function . has inflection points at (reading from left to right) , and where is and is and is