To determine the points of inflection, as well as the intervals where $f(x)$ is concave up and concave down, we need to compute the second derivative of $f(x)$ and analyze its sign.

Step 1: Given Function

The function is: $f(x) = (x - 8)(1 - x^3)$

Expanding $f(x)$ will make differentiation easier.

Step 2: Expand $f(x)$

$f(x) = (x - 8)(1 - x^3) = x - x^4 - 8 + 8x^3$ Thus, $f(x) = -x^4 + 8x^3 + x - 8$

Step 3: First Derivative $f'(x)$

Differentiating with respect to $x$: $f'(x) = -4x^3 + 24x^2 + 1$

Step 4: Second Derivative $f''(x)$

Differentiating $f'(x)$ to find $f''(x)$: $f''(x) = -12x^2 + 48x$ or $f''(x) = -12x(x - 4)$

Step 5: Solve $f''(x) = 0$ for Inflection Points

Setting $f''(x) = 0$: $-12x(x - 4) = 0$ This gives: $x = 0 \quad \text{and} \quad x = 4$

These are the possible inflection points. To confirm, we need to check the sign changes of $f''(x)$ around these points.

Step 6: Test the Intervals for Concavity

1. Interval $(-\infty, 0)$: Choose $x = -1$: $f''(-1) = -12(-1)((-1) - 4) = -12 \cdot (-1) \cdot (-5) = -60 \quad (\text{negative, concave down})$

2. Interval $(0, 4)$: Choose $x = 2$: $f''(2) = -12(2)(2 - 4) = -12 \cdot 2 \cdot (-2) = 48 \quad (\text{positive, concave up})$

3. Interval $(4, \infty)$: Choose $x = 5$: $f''(5) = -12(5)(5 - 4) = -12 \cdot 5 \cdot 1 = -60 \quad (\text{negative, concave down})$

Step 7: Summary of Results

• Points of inflection: Points where $x = 0$ and $x = 4$. To find $y$-values:

  • At $x = 0$: $f(0) = (0 - 8)(1 - 0^3) = -8$ So, one point of inflection is $(0, -8)$.
  • At $x = 4$: $f(4) = (4 - 8)(1 - 4^3) = -4 \cdot (-63) = 252$ So, another point of inflection is $(4, 252)$.

• Concavity intervals:

  • Concave up on $(0, 4)$
  • Concave down on $(-\infty, 0) \cup (4, \infty)$

Final Answer:

• Points of inflection: $(x, y) = (0, -8), (4, 252)$
• Concave up when $x \in (0, 4)$
• Concave down when $x \in (-\infty, 0) \cup (4, \infty)$

Would you like a breakdown of each calculation step, or do you have further questions?

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Follow-Up Questions:

1. How do you find concave up and down intervals using the second derivative?
2. Why are points where the second derivative is zero considered for inflection points?
3. Can a point where the second derivative is zero not be an inflection point? Why?
4. How does expanding $f(x)$ make differentiation easier in this example?
5. What other methods could we use to analyze concavity and inflection points?

Tip:

When solving for concavity, always check the sign changes of the second derivative around critical points, as they confirm true inflection points.