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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]
[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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