Question

Let $f(x) = (x-3)^5(x+1)^4$ then:

• A. $x = \frac{7}{9}$ is a point of maxima
• B. $x = 3$ is a point of minimum
• C. $x = -1$ is a point of maxima
• D. $f$ has no point of maximum or minimum

Answer 1

Answer: (C)

$x = -1$ is a point of maxima

Answer 2

To find the points of local maxima and minima for the function $f(x) = (x-3)^5(x+1)^4$, we first need to find its first derivative, $f'(x)$, and then determine the critical points by setting $f'(x) = 0$.

Step 1: Find the first derivative $f'(x)$.

Using the product rule $(uv)' = u'v + uv'$, where $u = (x-3)^5$ and $v = (x+1)^4$: $u' = \frac{d}{dx}(x-3)^5 = 5(x-3)^{5-1} \cdot \frac{d}{dx}(x-3) = 5(x-3)^4 \cdot 1 = 5(x-3)^4$ $v' = \frac{d}{dx}(x+1)^4 = 4(x+1)^{4-1} \cdot \frac{d}{dx}(x+1) = 4(x+1)^3 \cdot 1 = 4(x+1)^3$

Now, substitute these into the product rule formula: $f'(x) = u'v + uv'$ $f'(x) = 5(x-3)^4(x+1)^4 + (x-3)^5 \cdot 4(x+1)^3$

Factor out the common terms, which are $(x-3)^4$ and $(x+1)^3$: $f'(x) = (x-3)^4(x+1)^3 [5(x+1) + 4(x-3)]$ $f'(x) = (x-3)^4(x+1)^3 [5x + 5 + 4x - 12]$ $f'(x) = (x-3)^4(x+1)^3 [9x - 7]$

Step 2: Find the critical points.

Set $f'(x) = 0$ to find the critical points: $(x-3)^4(x+1)^3(9x - 7) = 0$

This equation holds true if any of its factors are zero:

1. $x-3 = 0 \implies x = 3$
2. $x+1 = 0 \implies x = -1$
3. $9x-7 = 0 \implies 9x = 7 \implies x = \frac{7}{9}$

The critical points are $x = -1, x = \frac{7}{9}, x = 3$.

Step 3: Use the First Derivative Test to classify the critical points.

We analyze the sign of $f'(x)$ in intervals around each critical point. The term $(x-3)^4$ is always non-negative, so its sign does not affect the sign change of $f'(x)$ when passing through $x=3$. The sign changes depend on $(x+1)^3$ and $(9x-7)$.

Let's arrange the critical points in ascending order: $-1, \frac{7}{9}, 3$.

• Interval 1: $x < -1$ (e.g., $x = -2$)

  • $(x-3)^4 = (-2-3)^4 = (-5)^4 = 625$ (Positive)
  • $(x+1)^3 = (-2+1)^3 = (-1)^3 = -1$ (Negative)
  • $(9x-7) = 9(-2)-7 = -18-7 = -25$ (Negative)
  • $f'(x) = (\text{Positive}) \times (\text{Negative}) \times (\text{Negative}) = \text{Positive}$

  So, $f(x)$ is increasing for $x < -1$.

• Interval 2: $-1 < x < \frac{7}{9}$ (e.g., $x = 0$)

  • $(x-3)^4 = (0-3)^4 = (-3)^4 = 81$ (Positive)
  • $(x+1)^3 = (0+1)^3 = 1^3 = 1$ (Positive)
  • $(9x-7) = 9(0)-7 = -7$ (Negative)
  • $f'(x) = (\text{Positive}) \times (\text{Positive}) \times (\text{Negative}) = \text{Negative}$

  So, $f(x)$ is decreasing for $-1 < x < \frac{7}{9}$.

  Conclusion for $x = -1$: As $f'(x)$ changes from positive to negative at $x = -1$, $x = -1$ is a point of local maximum.

• Interval 3: $\frac{7}{9} < x < 3$ (e.g., $x = 1$)

  • $(x-3)^4 = (1-3)^4 = (-2)^4 = 16$ (Positive)
  • $(x+1)^3 = (1+1)^3 = 2^3 = 8$ (Positive)
  • $(9x-7) = 9(1)-7 = 2$ (Positive)
  • $f'(x) = (\text{Positive}) \times (\text{Positive}) \times (\text{Positive}) = \text{Positive}$

  So, $f(x)$ is increasing for $\frac{7}{9} < x < 3$.

  Conclusion for $x = \frac{7}{9}$: As $f'(x)$ changes from negative to positive at $x = \frac{7}{9}$, $x = \frac{7}{9}$ is a point of local minimum.

• Interval 4: $x > 3$ (e.g., $x = 4$)

  • $(x-3)^4 = (4-3)^4 = 1^4 = 1$ (Positive)
  • $(x+1)^3 = (4+1)^3 = 5^3 = 125$ (Positive)
  • $(9x-7) = 9(4)-7 = 36-7 = 29$ (Positive)
  • $f'(x) = (\text{Positive}) \times (\text{Positive}) \times (\text{Positive}) = \text{Positive}$

  So, $f(x)$ is increasing for $x > 3$.

  Conclusion for $x = 3$: As $f'(x)$ does not change sign (it remains positive) at $x = 3$, $x = 3$ is a point of inflection, not a local maximum or minimum.

Step 4: Evaluate the given options.

1. $x = \frac{7}{9}$ is a point of maxima. (Incorrect, it's a minimum)
2. $x = 3$ is a point of minimum. (Incorrect, it's an inflection point)
3. $x = -1$ is a point of maxima. (Correct)
4. $f$ has no point of maximum or minimum. (Incorrect, it has a maximum and a minimum)

The only correct option is $x = -1$ is a point of maxima.