Answer :
- Find the second derivative of the function: $f''(x) = 240x^3 + 720x^2 - 1440x$.
- Set the second derivative equal to zero and solve for $x$ to find potential inflection points: $240x(x^2 + 3x - 6) = 0$, which gives $x = 0, \frac{-3 - \sqrt{33}}{2}, \frac{-3 + \sqrt{33}}{2}$.
- Determine the intervals based on the inflection points and test the sign of the second derivative in each interval to determine concavity.
- Conclude the concavity on each interval: $(-\infty, D)$: Concave Down, $(D, E)$: Concave Up, $(E, F)$: Concave Down, $(F, \infty)$: Concave Up, where $D = \frac{-3 - \sqrt{33}}{2}$, $E = 0$, and $F = \frac{-3 + \sqrt{33}}{2}$.
### Explanation
1. Problem Analysis
We are given the function $f(x) = 12x^5 + 60x^4 - 240x^3 + 1$. We need to find the inflection points and determine the concavity of the function on the given intervals.
2. First Derivative
First, we find the first derivative of $f(x)$:
$$f'(x) = 60x^4 + 240x^3 - 720x^2$$
3. Second Derivative
Next, we find the second derivative of $f(x)$:
$$f''(x) = 240x^3 + 720x^2 - 1440x$$
4. Finding Inflection Points
To find the inflection points, we set $f''(x) = 0$ and solve for $x$:
$$240x^3 + 720x^2 - 1440x = 0$$
$$240x(x^2 + 3x - 6) = 0$$
So, $x = 0$ is one solution. Now we solve the quadratic equation $x^2 + 3x - 6 = 0$ 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 other two solutions are $x = \frac{-3 - \sqrt{33}}{2} \approx -4.37$ and $x = \frac{-3 + \sqrt{33}}{2} \approx 1.37$.
5. Identifying D, E, and F
The inflection points are $x = \frac{-3 - \sqrt{33}}{2}$, $x = 0$, and $x = \frac{-3 + \sqrt{33}}{2}$. So, $D = \frac{-3 - \sqrt{33}}{2} \approx -4.37$, $E = 0$, and $F = \frac{-3 + \sqrt{33}}{2} \approx 1.37$.
6. Determining Concavity
Now we determine the concavity of $f(x)$ on the intervals $(-\infty, D)$, $(D, E)$, $(E, F)$, and $(F, \infty)$ by analyzing the sign of $f''(x)$ on each interval.
1. $(-\infty, D)$: Let's test $x = -5$. $f''(-5) = 240(-5)^3 + 720(-5)^2 - 1440(-5) = -30000 + 18000 + 7200 = -4800 < 0$. So, $f(x)$ is concave down on $(-\infty, D)$.
2. $(D, E)$: Let's test $x = -1$. $f''(-1) = 240(-1)^3 + 720(-1)^2 - 1440(-1) = -240 + 720 + 1440 = 1920 > 0$. So, $f(x)$ is concave up on $(D, E)$.
3. $(E, F)$: Let's test $x = 1$. $f''(1) = 240(1)^3 + 720(1)^2 - 1440(1) = 240 + 720 - 1440 = -480 < 0$. So, $f(x)$ is concave down on $(E, F)$.
4. $(F, \infty)$: Let's test $x = 2$. $f''(2) = 240(2)^3 + 720(2)^2 - 1440(2) = 1920 + 2880 - 2880 = 1920 > 0$. So, $f(x)$ is concave up on $(F, \infty)$.
7. Final Answer
Therefore, $D = \frac{-3 - \sqrt{33}}{2}$, $E = 0$, and $F = \frac{-3 + \sqrt{33}}{2}$.
* $(-\infty, D)$: Concave Down
* $(D, E)$: Concave Up
* $(E, F)$: Concave Down
* $(F, \infty)$: Concave Up
### Examples
Understanding concavity and inflection points is crucial in various real-world applications. For instance, when designing a bridge, engineers need to analyze the concavity of the structure to ensure its stability and safety. Similarly, in economics, understanding the concavity of a cost function can help businesses optimize their production levels to minimize costs.
- Set the second derivative equal to zero and solve for $x$ to find potential inflection points: $240x(x^2 + 3x - 6) = 0$, which gives $x = 0, \frac{-3 - \sqrt{33}}{2}, \frac{-3 + \sqrt{33}}{2}$.
- Determine the intervals based on the inflection points and test the sign of the second derivative in each interval to determine concavity.
- Conclude the concavity on each interval: $(-\infty, D)$: Concave Down, $(D, E)$: Concave Up, $(E, F)$: Concave Down, $(F, \infty)$: Concave Up, where $D = \frac{-3 - \sqrt{33}}{2}$, $E = 0$, and $F = \frac{-3 + \sqrt{33}}{2}$.
### Explanation
1. Problem Analysis
We are given the function $f(x) = 12x^5 + 60x^4 - 240x^3 + 1$. We need to find the inflection points and determine the concavity of the function on the given intervals.
2. First Derivative
First, we find the first derivative of $f(x)$:
$$f'(x) = 60x^4 + 240x^3 - 720x^2$$
3. Second Derivative
Next, we find the second derivative of $f(x)$:
$$f''(x) = 240x^3 + 720x^2 - 1440x$$
4. Finding Inflection Points
To find the inflection points, we set $f''(x) = 0$ and solve for $x$:
$$240x^3 + 720x^2 - 1440x = 0$$
$$240x(x^2 + 3x - 6) = 0$$
So, $x = 0$ is one solution. Now we solve the quadratic equation $x^2 + 3x - 6 = 0$ 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 other two solutions are $x = \frac{-3 - \sqrt{33}}{2} \approx -4.37$ and $x = \frac{-3 + \sqrt{33}}{2} \approx 1.37$.
5. Identifying D, E, and F
The inflection points are $x = \frac{-3 - \sqrt{33}}{2}$, $x = 0$, and $x = \frac{-3 + \sqrt{33}}{2}$. So, $D = \frac{-3 - \sqrt{33}}{2} \approx -4.37$, $E = 0$, and $F = \frac{-3 + \sqrt{33}}{2} \approx 1.37$.
6. Determining Concavity
Now we determine the concavity of $f(x)$ on the intervals $(-\infty, D)$, $(D, E)$, $(E, F)$, and $(F, \infty)$ by analyzing the sign of $f''(x)$ on each interval.
1. $(-\infty, D)$: Let's test $x = -5$. $f''(-5) = 240(-5)^3 + 720(-5)^2 - 1440(-5) = -30000 + 18000 + 7200 = -4800 < 0$. So, $f(x)$ is concave down on $(-\infty, D)$.
2. $(D, E)$: Let's test $x = -1$. $f''(-1) = 240(-1)^3 + 720(-1)^2 - 1440(-1) = -240 + 720 + 1440 = 1920 > 0$. So, $f(x)$ is concave up on $(D, E)$.
3. $(E, F)$: Let's test $x = 1$. $f''(1) = 240(1)^3 + 720(1)^2 - 1440(1) = 240 + 720 - 1440 = -480 < 0$. So, $f(x)$ is concave down on $(E, F)$.
4. $(F, \infty)$: Let's test $x = 2$. $f''(2) = 240(2)^3 + 720(2)^2 - 1440(2) = 1920 + 2880 - 2880 = 1920 > 0$. So, $f(x)$ is concave up on $(F, \infty)$.
7. Final Answer
Therefore, $D = \frac{-3 - \sqrt{33}}{2}$, $E = 0$, and $F = \frac{-3 + \sqrt{33}}{2}$.
* $(-\infty, D)$: Concave Down
* $(D, E)$: Concave Up
* $(E, F)$: Concave Down
* $(F, \infty)$: Concave Up
### Examples
Understanding concavity and inflection points is crucial in various real-world applications. For instance, when designing a bridge, engineers need to analyze the concavity of the structure to ensure its stability and safety. Similarly, in economics, understanding the concavity of a cost function can help businesses optimize their production levels to minimize costs.
