Consider the function
f ( x )

To analyze the inflection points and concavity of the function $f(x) = 12x^5 + 30x^4 - 300x^3 + 6$, we need to follow these steps:

1. Find the second derivative $f''(x)$ of $f(x)$: Inflection points occur where the second derivative changes sign, which is typically where $f''(x) = 0$ or $f''(x)$ is undefined.

2. Solve for $f''(x) = 0$ to find potential inflection points.

3. Determine concavity on intervals defined by these inflection points by testing the sign of $f''(x)$ in each interval.

Let's go through these calculations:

Step 1: Find $f'(x)$

First, compute the first derivative $f'(x)$ of $f(x) = 12x^5 + 30x^4 - 300x^3 + 6$: $f'(x) = 60x^4 + 120x^3 - 900x^2$

Step 2: Find $f''(x)$

Now, compute the second derivative: $f''(x) = 240x^3 + 360x^2 - 1800x$ This simplifies to: $f''(x) = 60x(4x^2 + 6x - 30)$

Step 3: Solve $f''(x) = 0$

To find inflection points, set $f''(x) = 0$: $60x(4x^2 + 6x - 30) = 0$

This gives two factors:

1. $60x = 0$ ⟹ $x = 0$
2. $4x^2 + 6x - 30 = 0$

Solving $4x^2 + 6x - 30 = 0$ using the quadratic formula: $x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 4 \cdot (-30)}}{2 \cdot 4}$ $x = \frac{-6 \pm \sqrt{36 + 480}}{8}$ $x = \frac{-6 \pm \sqrt{516}}{8}$ $x = \frac{-6 \pm 22.7}{8}$ Thus, the solutions are: $x = 2.0875 \quad \text{and} \quad x = -3.5875$

So, the inflection points are approximately:

• $D = -3.5875$
• $E = 0$
• $F = 2.0875$

Step 4: Determine Concavity on Each Interval

Now, let's test the sign of $f''(x)$ on the intervals $(-\infty, D)$, $(D, E)$, $(E, F)$, and $(F, \infty)$:

1. On $(-\infty, D)$: Choose a test point $x < -3.5875$ (e.g., $x = -4$). Plugging into $f''(x)$, we find that $f''(x) > 0$, so $f(x)$ is concave up on $(-\infty, D)$.

2. On $(D, E)$: Choose a test point between $-3.5875$ and $0$ (e.g., $x = -2$). Plugging into $f''(x)$, we find that $f''(x) < 0$, so $f(x)$ is concave down on $(D, E)$.

3. On $(E, F)$: Choose a test point between $0$ and $2.0875$ (e.g., $x = 1$). Plugging into $f''(x)$, we find that $f''(x) > 0$, so $f(x)$ is concave up on $(E, F)$.

4. On $(F, \infty)$: Choose a test point $x > 2.0875$ (e.g., $x = 3$). Plugging into $f''(x)$, we find that $f''(x) < 0$, so $f(x)$ is concave down on $(F, \infty)$.

Summary of Answers

• $(- \infty, D)$: Concave up
• $(D, E)$: Concave down
• $(E, F)$: Concave up
• $(F, \infty)$: Concave down

Would you like more details on any of these steps or help with related problems?

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1. What are the implications of an inflection point for the function's graph?
2. How can you find inflection points using just the first derivative?
3. What is the difference between concavity and convexity?
4. How would concavity affect optimization problems?
5. Can a function have an inflection point where the second derivative is not zero?

Tip: To confirm concavity, always test points within intervals between critical points to verify the behavior.