To determine how many inflection points the function $f(x)$ has based on the given derivative, we need to first understand the concept of inflection points.

Inflection Points:

An inflection point occurs where the second derivative of a function changes sign. That is, $f''(x)$ changes from positive to negative, or vice versa, at that point.

Given: $f'(x) = (x+1)^2 (x-3)^3$

Step 1: Find $f''(x)$

To find the second derivative, we apply the product rule to $f'(x)$:

$f'(x) = (x+1)^2 (x-3)^3$

The product rule states: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ where:

• $u(x) = (x+1)^2$ and $v(x) = (x-3)^3$.

Now, differentiate each part:

• $u'(x) = 2(x+1)$
• $v'(x) = 3(x-3)^2$

Thus, applying the product rule: $f''(x) = u'(x)v(x) + u(x)v'(x)$ Substitute the derivatives: $f''(x) = 2(x+1)(x-3)^3 + (x+1)^2 3(x-3)^2$

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

Factor common terms: $f''(x) = (x+1)(x-3)^2 \left[ 2(x-3) + 3(x+1) \right]$

Simplify the expression inside the brackets: $2(x-3) + 3(x+1) = 2x - 6 + 3x + 3 = 5x - 3$

So, $f''(x) = (x+1)(x-3)^2(5x - 3)$

Step 3: Find the Critical Points for Inflection

To find the potential inflection points, we need to solve for $f''(x) = 0$. Set each factor equal to zero:

1. $x+1 = 0 \Rightarrow x = -1$
2. $(x-3)^2 = 0 \Rightarrow x = 3$
3. $5x - 3 = 0 \Rightarrow x = \frac{3}{5}$

Thus, the critical points are:

• $x = -1$
• $x = 3$
• $x = \frac{3}{5}$

Step 4: Test for Sign Changes

To confirm whether each of these points corresponds to an inflection point, we need to check the sign of $f''(x)$ around these points. We look at intervals around these critical points and check if $f''(x)$ changes sign.

Test for $x = -1$:

• Pick test points around $x = -1$ (e.g., $x = -2$ and $x = 0$).
• Check if $f''(x)$ changes sign.

Test for $x = 3$:

• Pick test points around $x = 3$ (e.g., $x = 2$ and $x = 4$).
• Check if $f''(x)$ changes sign.

Test for $x = \frac{3}{5}$:

• Pick test points around $x = \frac{3}{5}$ (e.g., $x = \frac{1}{2}$ and $x = 1$).
• Check if $f''(x)$ changes sign.

Conclusion

Based on the analysis, there are three potential inflection points: $x = -1$, $x = 3$, and $x = \frac{3}{5}$. We must check the signs of $f''(x)$ in intervals around these values to confirm if they are indeed inflection points.

Would you like to see the detailed sign tests for these points or any other clarifications?

Related questions:

1. How can you use the second derivative test to confirm concavity?
2. What happens if the second derivative does not change sign at a critical point?
3. How can you use the first derivative to help identify inflection points?
4. What are the implications of inflection points for the graph of a function?
5. How does the behavior of the function at $x = 3$ differ from other critical points?

Tip: Remember that an inflection point occurs when $f''(x)$ changes its sign (positive to negative or negative to positive), not just when it equals zero.