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Consider the function [tex]f(x) = 12x^5 + 60x^4 - 100x^3 + 1[/tex].

[tex]f(x)[/tex] has inflection points at (reading from left to right) [tex]x = D, E[/tex], and [tex]F[/tex], where:
- [tex]D[/tex] is [tex]\square[/tex]
- [tex]E[/tex] is [tex]\square[/tex]
- [tex]F[/tex] is [tex]\square[/tex]

For each of the following intervals, determine whether [tex]f(x)[/tex] is concave up or concave down.

\[
\begin{array}{ll}
(-\infty, D): & \text{Select an answer} \\
(D, E): & \text{Select an answer} \\
(E, F): & \text{Select an answer} \\
(F, \infty): & \text{Select an answer}
\end{array}
\]

Answer :

We start with the function

[tex]$$
f(x) = 12x^5 + 60x^4 - 100x^3 + 1.
$$[/tex]

### Step 1. Find the Inflection Points

An inflection point occurs where the concavity changes, that is, where the second derivative changes sign. We first compute the second derivative.

#### First, compute the first derivative:

[tex]$$
f'(x) = 60x^4 + 240x^3 - 300x^2.
$$[/tex]

#### Next, compute the second derivative:

A calculation shows that

[tex]$$
f''(x) = 120x\,(2x^2 + 6x - 5).
$$[/tex]

Setting the second derivative equal to zero gives

[tex]$$
120x\,(2x^2 + 6x - 5) = 0.
$$[/tex]

This equation is satisfied when

1. [tex]$$120x = 0 \quad \Longrightarrow \quad x = 0,$$[/tex]

2. [tex]$$2x^2 + 6x - 5 = 0.$$[/tex]

For the quadratic equation, apply the quadratic formula:

[tex]$$
x = \frac{-6 \pm \sqrt{6^2 - 4(2)(-5)}}{2 \cdot 2} = \frac{-6 \pm \sqrt{36 + 40}}{4} = \frac{-6 \pm \sqrt{76}}{4}.
$$[/tex]

Notice that

[tex]$$
\sqrt{76} = 2\sqrt{19},
$$[/tex]

so

[tex]$$
x = \frac{-6 \pm 2\sqrt{19}}{4} = \frac{-3 \pm \sqrt{19}}{2}.
$$[/tex]

Thus, the three inflection points are given by

[tex]\[
D = \frac{-3 - \sqrt{19}}{2}, \quad E = 0, \quad F = \frac{-3 + \sqrt{19}}{2}.
\][/tex]

Numerically, these are approximately

[tex]\[
D \approx -3.68, \quad E = 0, \quad F \approx 0.68.
\][/tex]

### Step 2. Determine the Concavity on Each Interval

The sign of [tex]$f''(x)$[/tex] will tell us the concavity:

- If [tex]$f''(x) > 0$[/tex], [tex]$f(x)$[/tex] is concave up.
- If [tex]$f''(x) < 0$[/tex], [tex]$f(x)$[/tex] is concave down.

We test a sample point in each interval determined by the inflection points.

#### 1. Interval [tex]$(-\infty, D)$[/tex]

Choose a test point less than [tex]$D$[/tex], for instance, a point slightly less than [tex]$-3.68$[/tex]. When computed, [tex]$f''(x)$[/tex] is negative in this region. Therefore,

[tex]$$
(-\infty, D): \text{concave down}.
$$[/tex]

#### 2. Interval [tex]$(D, E)$[/tex]

Pick a test point between [tex]$D \approx -3.68$[/tex] and [tex]$E = 0$[/tex], say around [tex]$-1.84$[/tex]. In this interval, the calculation shows that [tex]$f''(x)$[/tex] is positive. Hence,

[tex]$$
(D, E): \text{concave up}.
$$[/tex]

#### 3. Interval [tex]$(E, F)$[/tex]

Choose a test point between [tex]$E = 0$[/tex] and [tex]$F \approx 0.68$[/tex], such as [tex]$0.34$[/tex]. Here, the value of [tex]$f''(x)$[/tex] turns out to be negative. So,

[tex]$$
(E, F): \text{concave down}.
$$[/tex]

#### 4. Interval [tex]$(F, \infty)$[/tex]

Take a point greater than [tex]$F$[/tex], for example, slightly more than [tex]$0.68$[/tex]. The second derivative is positive, so

[tex]$$
(F, \infty): \text{concave up}.
$$[/tex]

### Final Answer

- The inflection points occur at
[tex]$$ D \approx -3.68, \quad E = 0, \quad F \approx 0.68. $$[/tex]
- The concavity on each interval is:
[tex]$$ (-\infty, D): \text{concave down}, $$[/tex]
[tex]$$ (D, E): \text{concave up}, $$[/tex]
[tex]$$ (E, F): \text{concave down}, $$[/tex]
[tex]$$ (F, \infty): \text{concave up}. $$[/tex]

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