Answer :
We begin with the function
[tex]$$
f(x) = 12x^5 + 45x^4 - 360x^3 + 3.
$$[/tex]
Step 1. Find the second derivative.
First, compute the first derivative:
[tex]$$
f'(x) = \frac{d}{dx}(12x^5) + \frac{d}{dx}(45x^4) - \frac{d}{dx}(360x^3) + \frac{d}{dx}(3) = 60x^4 + 180x^3 - 1080x^2.
$$[/tex]
Now, differentiate again to obtain the second derivative:
[tex]$$
f''(x) = \frac{d}{dx}(60x^4) + \frac{d}{dx}(180x^3) - \frac{d}{dx}(1080x^2) = 240x^3 + 540x^2 - 2160x.
$$[/tex]
Step 2. Factor the second derivative to find the potential inflection points.
Factor out the common term [tex]$60x$[/tex]:
[tex]$$
f''(x) = 60x\left(4x^2 + 9x - 36\right).
$$[/tex]
Set [tex]$f''(x) = 0$[/tex]:
[tex]$$
60x\left(4x^2 + 9x - 36\right) = 0.
$$[/tex]
This equation is satisfied when
1. [tex]$$60x = 0 \quad \Longrightarrow \quad x=0,$$[/tex]
2. or when
[tex]$$
4x^2 + 9x - 36 = 0.
$$[/tex]
Solve the quadratic equation with the quadratic formula:
[tex]$$
x = \frac{-9 \pm \sqrt{9^2 - 4\cdot4\cdot(-36)}}{2\cdot4} = \frac{-9 \pm \sqrt{81 + 576}}{8} = \frac{-9 \pm \sqrt{657}}{8}.
$$[/tex]
Notice that [tex]$\sqrt{657}$[/tex] can be expressed as [tex]$3\sqrt{73}$[/tex] because [tex]$657 = 9 \cdot 73$[/tex]. Hence, the two solutions are:
[tex]$$
x = \frac{-9 - 3\sqrt{73}}{8} \quad \text{and} \quad x = \frac{-9 + 3\sqrt{73}}{8}.
$$[/tex]
Step 3. Identify the inflection points.
Label the inflection points in order from left to right as [tex]$D$[/tex], [tex]$E$[/tex], and [tex]$F$[/tex]. Comparing the three roots:
- The smallest (leftmost) is
[tex]$$
D = \frac{-9 - 3\sqrt{73}}{8}.
$$[/tex]
- The middle value is
[tex]$$
E = 0.
$$[/tex]
- The largest (rightmost) is
[tex]$$
F = \frac{-9 + 3\sqrt{73}}{8}.
$$[/tex]
For approximation purposes, we have:
[tex]$$
D \approx -4.329,\quad E = 0,\quad F \approx 2.079.
$$[/tex]
Step 4. Determine the concavity on each interval.
The sign of [tex]$f''(x)$[/tex] tells us the concavity:
- If [tex]$f''(x) > 0$[/tex] then [tex]$f(x)$[/tex] is concave up.
- If [tex]$f''(x) < 0$[/tex] then [tex]$f(x)$[/tex] is concave down.
Examine the four intervals determined by [tex]$D$[/tex], [tex]$E$[/tex], and [tex]$F$[/tex]:
1. For the interval [tex]$(-\infty, D)$[/tex] (values less than approximately [tex]$-4.329$[/tex]), choose a test point (for example, [tex]$x = -5$[/tex]). The evaluation shows [tex]$f''(x) < 0$[/tex], meaning the graph is concave down on [tex]$(-\infty, D)$[/tex].
2. For the interval [tex]$(D, E)$[/tex] (values between approximately [tex]$-4.329$[/tex] and [tex]$0$[/tex]), choose a test point (for example, [tex]$x = -2$[/tex]). The evaluation shows [tex]$f''(x) > 0$[/tex], meaning the graph is concave up on [tex]$(D, E)$[/tex].
3. For the interval [tex]$(E, F)$[/tex] (values between [tex]$0$[/tex] and approximately [tex]$2.079$[/tex]), choose a test point (for example, [tex]$x = 1$[/tex]). The evaluation shows [tex]$f''(x) < 0$[/tex], meaning the graph is concave down on [tex]$(E, F)$[/tex].
4. For the interval [tex]$(F, \infty)$[/tex] (values greater than approximately [tex]$2.079$[/tex]), choose a test point (for example, [tex]$x = 3$[/tex]). The evaluation shows [tex]$f''(x) > 0$[/tex], meaning the graph is concave up on [tex]$(F, \infty)$[/tex].
Final Answers:
Inflection points:
[tex]$$
D = \frac{-9 - 3\sqrt{73}}{8},\quad E = 0,\quad F = \frac{-9 + 3\sqrt{73}}{8}.
$$[/tex]
Concavity:
- On [tex]$(-\infty, D)$[/tex]: Concave Down.
- On [tex]$(D, E)$[/tex]: Concave Up.
- On [tex]$(E, F)$[/tex]: Concave Down.
- On [tex]$(F, \infty)$[/tex]: Concave Up.
[tex]$$
f(x) = 12x^5 + 45x^4 - 360x^3 + 3.
$$[/tex]
Step 1. Find the second derivative.
First, compute the first derivative:
[tex]$$
f'(x) = \frac{d}{dx}(12x^5) + \frac{d}{dx}(45x^4) - \frac{d}{dx}(360x^3) + \frac{d}{dx}(3) = 60x^4 + 180x^3 - 1080x^2.
$$[/tex]
Now, differentiate again to obtain the second derivative:
[tex]$$
f''(x) = \frac{d}{dx}(60x^4) + \frac{d}{dx}(180x^3) - \frac{d}{dx}(1080x^2) = 240x^3 + 540x^2 - 2160x.
$$[/tex]
Step 2. Factor the second derivative to find the potential inflection points.
Factor out the common term [tex]$60x$[/tex]:
[tex]$$
f''(x) = 60x\left(4x^2 + 9x - 36\right).
$$[/tex]
Set [tex]$f''(x) = 0$[/tex]:
[tex]$$
60x\left(4x^2 + 9x - 36\right) = 0.
$$[/tex]
This equation is satisfied when
1. [tex]$$60x = 0 \quad \Longrightarrow \quad x=0,$$[/tex]
2. or when
[tex]$$
4x^2 + 9x - 36 = 0.
$$[/tex]
Solve the quadratic equation with the quadratic formula:
[tex]$$
x = \frac{-9 \pm \sqrt{9^2 - 4\cdot4\cdot(-36)}}{2\cdot4} = \frac{-9 \pm \sqrt{81 + 576}}{8} = \frac{-9 \pm \sqrt{657}}{8}.
$$[/tex]
Notice that [tex]$\sqrt{657}$[/tex] can be expressed as [tex]$3\sqrt{73}$[/tex] because [tex]$657 = 9 \cdot 73$[/tex]. Hence, the two solutions are:
[tex]$$
x = \frac{-9 - 3\sqrt{73}}{8} \quad \text{and} \quad x = \frac{-9 + 3\sqrt{73}}{8}.
$$[/tex]
Step 3. Identify the inflection points.
Label the inflection points in order from left to right as [tex]$D$[/tex], [tex]$E$[/tex], and [tex]$F$[/tex]. Comparing the three roots:
- The smallest (leftmost) is
[tex]$$
D = \frac{-9 - 3\sqrt{73}}{8}.
$$[/tex]
- The middle value is
[tex]$$
E = 0.
$$[/tex]
- The largest (rightmost) is
[tex]$$
F = \frac{-9 + 3\sqrt{73}}{8}.
$$[/tex]
For approximation purposes, we have:
[tex]$$
D \approx -4.329,\quad E = 0,\quad F \approx 2.079.
$$[/tex]
Step 4. Determine the concavity on each interval.
The sign of [tex]$f''(x)$[/tex] tells us the concavity:
- If [tex]$f''(x) > 0$[/tex] then [tex]$f(x)$[/tex] is concave up.
- If [tex]$f''(x) < 0$[/tex] then [tex]$f(x)$[/tex] is concave down.
Examine the four intervals determined by [tex]$D$[/tex], [tex]$E$[/tex], and [tex]$F$[/tex]:
1. For the interval [tex]$(-\infty, D)$[/tex] (values less than approximately [tex]$-4.329$[/tex]), choose a test point (for example, [tex]$x = -5$[/tex]). The evaluation shows [tex]$f''(x) < 0$[/tex], meaning the graph is concave down on [tex]$(-\infty, D)$[/tex].
2. For the interval [tex]$(D, E)$[/tex] (values between approximately [tex]$-4.329$[/tex] and [tex]$0$[/tex]), choose a test point (for example, [tex]$x = -2$[/tex]). The evaluation shows [tex]$f''(x) > 0$[/tex], meaning the graph is concave up on [tex]$(D, E)$[/tex].
3. For the interval [tex]$(E, F)$[/tex] (values between [tex]$0$[/tex] and approximately [tex]$2.079$[/tex]), choose a test point (for example, [tex]$x = 1$[/tex]). The evaluation shows [tex]$f''(x) < 0$[/tex], meaning the graph is concave down on [tex]$(E, F)$[/tex].
4. For the interval [tex]$(F, \infty)$[/tex] (values greater than approximately [tex]$2.079$[/tex]), choose a test point (for example, [tex]$x = 3$[/tex]). The evaluation shows [tex]$f''(x) > 0$[/tex], meaning the graph is concave up on [tex]$(F, \infty)$[/tex].
Final Answers:
Inflection points:
[tex]$$
D = \frac{-9 - 3\sqrt{73}}{8},\quad E = 0,\quad F = \frac{-9 + 3\sqrt{73}}{8}.
$$[/tex]
Concavity:
- On [tex]$(-\infty, D)$[/tex]: Concave Down.
- On [tex]$(D, E)$[/tex]: Concave Up.
- On [tex]$(E, F)$[/tex]: Concave Down.
- On [tex]$(F, \infty)$[/tex]: Concave Up.
