Key Concepts and Formulas

• First Derivative Test: A method for finding local maxima and minima of a function by analyzing the sign changes of its first derivative.
• Product Rule: $\frac{d}{dx}(uv) = u'v + uv'$, where $u$ and $v$ are functions of $x$.
• Critical Points: Points where the first derivative is either zero or undefined.

Step-by-Step Solution

Step 1: Find the First Derivative of f(x)

We are given $f(x) = (x-3)^{n_1}(x-5)^{n_2}$. We need to find $f'(x)$ using the product rule. $f'(x) = \frac{d}{dx} \left[ (x-3)^{n_1} (x-5)^{n_2} \right]$ Applying the product rule, we get: $f'(x) = n_1(x-3)^{n_1-1}(x-5)^{n_2} + (x-3)^{n_1}n_2(x-5)^{n_2-1}$ Now, we factor out the common terms $(x-3)^{n_1-1}$ and $(x-5)^{n_2-1}$: $f'(x) = (x-3)^{n_1-1}(x-5)^{n_2-1}[n_1(x-5) + n_2(x-3)]$ This is the simplified form of the first derivative.

Step 2: Find the Critical Points

Critical points occur when $f'(x) = 0$. Thus, $(x-3)^{n_1-1}(x-5)^{n_2-1}[n_1(x-5) + n_2(x-3)] = 0$ This gives us three possible critical points:

1. $x-3 = 0 \implies x = 3$
2. $x-5 = 0 \implies x = 5$
3. $n_1(x-5) + n_2(x-3) = 0 \implies n_1x - 5n_1 + n_2x - 3n_2 = 0 \implies (n_1+n_2)x = 5n_1 + 3n_2 \implies x = \frac{5n_1 + 3n_2}{n_1+n_2}$ Let $\alpha = \frac{5n_1 + 3n_2}{n_1+n_2}$. We are interested in the interval $(3, 5)$, so we need to check if $\alpha \in (3, 5)$.

Step 3: Analyze Option A: $n_1 = 3, n_2 = 4$

Substitute $n_1 = 3$ and $n_2 = 4$ into the expression for $\alpha$: $\alpha = \frac{5(3) + 3(4)}{3+4} = \frac{15 + 12}{7} = \frac{27}{7} \approx 3.857$ Since $3 < \frac{27}{7} < 5$, $\alpha \in (3, 5)$. Now we analyze the sign of $f'(x)$ around $\alpha$. $f'(x) = (x-3)^{3-1}(x-5)^{4-1}[3(x-5) + 4(x-3)] = (x-3)^2(x-5)^3(7x-27)$

• For $x \in (3, \frac{27}{7})$, $(x-3)^2 > 0$, $(x-5)^3 < 0$, $(7x-27) < 0$. Thus, $f'(x) = (+)(-)(-) = +$.
• For $x \in (\frac{27}{7}, 5)$, $(x-3)^2 > 0$, $(x-5)^3 < 0$, $(7x-27) > 0$. Thus, $f'(x) = (+)(-)(+) = -$. Since $f'(x)$ changes from positive to negative at $x = \frac{27}{7}$, there is a local maximum at $x = \frac{27}{7}$. Therefore, option (A) is TRUE.

Step 4: Analyze Option B: $n_1 = 4, n_2 = 3$

Substitute $n_1 = 4$ and $n_2 = 3$ into the expression for $\alpha$: $\alpha = \frac{5(4) + 3(3)}{4+3} = \frac{20 + 9}{7} = \frac{29}{7} \approx 4.143$ Since $3 < \frac{29}{7} < 5$, $\alpha \in (3, 5)$. Now we analyze the sign of $f'(x)$ around $\alpha$. $f'(x) = (x-3)^{4-1}(x-5)^{3-1}[4(x-5) + 3(x-3)] = (x-3)^3(x-5)^2(7x-29)$

• For $x \in (3, \frac{29}{7})$, $(x-3)^3 > 0$, $(x-5)^2 > 0$, $(7x-29) < 0$. Thus, $f'(x) = (+)(+)(-) = -$.
• For $x \in (\frac{29}{7}, 5)$, $(x-3)^3 > 0$, $(x-5)^2 > 0$, $(7x-29) > 0$. Thus, $f'(x) = (+)(+)(+) = +$. Since $f'(x)$ changes from negative to positive at $x = \frac{29}{7}$, there is a local minimum at $x = \frac{29}{7}$. Therefore, option (B) is TRUE.

Step 5: Analyze Option C: $n_1 = 3, n_2 = 5$

Substitute $n_1 = 3$ and $n_2 = 5$ into the expression for $\alpha$: $\alpha = \frac{5(3) + 3(5)}{3+5} = \frac{15 + 15}{8} = \frac{30}{8} = \frac{15}{4} = 3.75$ Since $3 < \frac{15}{4} < 5$, $\alpha \in (3, 5)$. Now we analyze the sign of $f'(x)$ around $\alpha$. $f'(x) = (x-3)^{3-1}(x-5)^{5-1}[3(x-5) + 5(x-3)] = (x-3)^2(x-5)^4(8x-30) = 2(x-3)^2(x-5)^4(4x-15)$

• For $x \in (3, \frac{15}{4})$, $(x-3)^2 > 0$, $(x-5)^4 > 0$, $(4x-15) < 0$. Thus, $f'(x) = (+)(+)(-) = -$.
• For $x \in (\frac{15}{4}, 5)$, $(x-3)^2 > 0$, $(x-5)^4 > 0$, $(4x-15) > 0$. Thus, $f'(x) = (+)(+)(+) = +$. Since $f'(x)$ changes from negative to positive at $x = \frac{15}{4}$, there is a local minimum at $x = \frac{15}{4}$. Therefore, there is NO local maxima, so option (C) is FALSE. However, we need to find the one that is NOT true.

Step 6: Analyze Option D: $n_1 = 4, n_2 = 6$

Substitute $n_1 = 4$ and $n_2 = 6$ into the expression for $\alpha$: $\alpha = \frac{5(4) + 3(6)}{4+6} = \frac{20 + 18}{10} = \frac{38}{10} = \frac{19}{5} = 3.8$ Since $3 < \frac{19}{5} < 5$, $\alpha \in (3, 5)$. Now we analyze the sign of $f'(x)$ around $\alpha$. $f'(x) = (x-3)^{4-1}(x-5)^{6-1}[4(x-5) + 6(x-3)] = (x-3)^3(x-5)^5(10x-38) = 2(x-3)^3(x-5)^5(5x-19)$

• For $x \in (3, \frac{19}{5})$, $(x-3)^3 > 0$, $(x-5)^5 < 0$, $(5x-19) < 0$. Thus, $f'(x) = (+)(-)(-) = +$.
• For $x \in (\frac{19}{5}, 5)$, $(x-3)^3 > 0$, $(x-5)^5 < 0$, $(5x-19) > 0$. Thus, $f'(x) = (+)(-)(+) = -$. Since $f'(x)$ changes from positive to negative at $x = \frac{19}{5}$, there is a local maximum at $x = \frac{19}{5}$. Therefore, option (D) is TRUE.

Since options B, C, and D are true, A must be incorrect.

Recalculating Option C: $f'(x) = (x-3)^2 (x-5)^4 (8x - 30) = 2(x-3)^2 (x-5)^4 (4x - 15)$ The critical point is $x = \frac{15}{4} = 3.75$. When $x < \frac{15}{4}$, $f'(x) < 0$. When $x > \frac{15}{4}$, $f'(x) > 0$. Thus, there is a local minimum. Option C states that there is a local maximum, which is incorrect.

Common Mistakes & Tips

• Remember to use the product rule correctly when finding the derivative.
• Pay attention to the exponents when determining the sign of the derivative around critical points.
• When analyzing the sign of $f'(x)$, consider the signs of each factor in the expression.

Summary

We analyzed the given options by finding the first derivative of the function, identifying the critical points, and examining the sign changes of the derivative around those points. We found that option C states that there is a local maxima, which is incorrect since there is a local minima.

Final Answer

The final answer is \boxed{C}, which corresponds to option (C).