Key Concepts and Formulas

• First Derivative Test for Local Extrema: If $f'(c) = 0$ or $f'(c)$ is undefined, then $x=c$ is a critical point. If $f'(x)$ changes sign from negative to positive at $x=c$, then $f(x)$ has a local minimum at $x=c$. If $f'(x)$ changes sign from positive to negative at $x=c$, then $f(x)$ has a local maximum at $x=c$.
• Second Derivative Test for Local Extrema: If $f'(c) = 0$ and $f''(c) > 0$, then $f(x)$ has a local minimum at $x=c$. If $f'(c) = 0$ and $f''(c) < 0$, then $f(x)$ has a local maximum at $x=c$. If $f''(c) = 0$, the test is inconclusive.
• Point of Inflection: A point $x=c$ is a point of inflection if $f''(c) = 0$ or $f''(c)$ is undefined, and $f''(x)$ changes sign at $x=c$.

Step-by-Step Solution

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

We are given $f(x) = 3^{(x^2 - 2)^3 + 4}$. We need to find $f'(x)$. Using the chain rule, we have:

$f'(x) = 3^{(x^2 - 2)^3 + 4} \cdot \ln(3) \cdot \frac{d}{dx} [(x^2 - 2)^3 + 4]$ $f'(x) = 3^{(x^2 - 2)^3 + 4} \cdot \ln(3) \cdot [3(x^2 - 2)^2 \cdot (2x)]$ $f'(x) = 6x(x^2 - 2)^2 \cdot \ln(3) \cdot 3^{(x^2 - 2)^3 + 4}$

Step 2: Find the critical points

To find the critical points, we set $f'(x) = 0$. Since $3^{(x^2 - 2)^3 + 4}$ and $\ln(3)$ are always positive, we only need to consider:

$6x(x^2 - 2)^2 = 0$ This gives us $x = 0$ and $x^2 - 2 = 0 \implies x = \pm \sqrt{2}$. Thus, the critical points are $x = -\sqrt{2}, 0, \sqrt{2}$.

Step 3: Analyze the sign of $f'(x)$ around $x=0$

Let's analyze the sign of $f'(x) = 6x(x^2 - 2)^2 \cdot \ln(3) \cdot 3^{(x^2 - 2)^3 + 4}$ around $x = 0$. Note that $(x^2 - 2)^2$ is always non-negative.

• For $x < 0$ (and close to 0), $6x < 0$, and $(x^2 - 2)^2 > 0$. Therefore, $f'(x) < 0$.
• For $x > 0$ (and close to 0), $6x > 0$, and $(x^2 - 2)^2 > 0$. Therefore, $f'(x) > 0$. Since $f'(x)$ changes from negative to positive at $x = 0$, $x = 0$ is a point of local minima. Thus, statement P is true.

Step 4: Find the second derivative of $f(x)$

To analyze points of inflection, we need to find $f''(x)$. Let $f'(x) = 6 \ln(3) x (x^2-2)^2 3^{(x^2-2)^3+4}$. Let $u(x) = x(x^2-2)^2$ and $v(x) = 3^{(x^2-2)^3+4}$. Then $f'(x) = 6\ln(3) u(x)v(x)$.

We already have $v'(x) = 6x(x^2-2)^2 \ln(3) 3^{(x^2-2)^3+4}$. Now, let's calculate $u'(x)$. $u(x) = x(x^4 - 4x^2 + 4) = x^5 - 4x^3 + 4x$ $u'(x) = 5x^4 - 12x^2 + 4$ So, $f''(x) = 6 \ln(3) [u'(x)v(x) + u(x)v'(x)]$ $f''(x) = 6 \ln(3) [(5x^4 - 12x^2 + 4)3^{(x^2-2)^3+4} + x(x^2-2)^2(6x(x^2-2)^2 \ln(3) 3^{(x^2-2)^3+4})]$ $f''(x) = 6 \ln(3) 3^{(x^2-2)^3+4} [(5x^4 - 12x^2 + 4) + 6x^2(x^2-2)^4 \ln(3)]$

Step 5: Check if $x=\sqrt{2}$ is a point of inflection.

We need to check if $f''(\sqrt{2}) = 0$ and if $f''(x)$ changes sign at $x = \sqrt{2}$. $f''(\sqrt{2}) = 6 \ln(3) 3^{(2-2)^3+4} [(5(4) - 12(2) + 4) + 6(2)(2-2)^4 \ln(3)] = 6 \ln(3) \cdot 3^4 (20 - 24 + 4) = 6 \ln(3) \cdot 3^4 \cdot 0 = 0$. Now, we need to check if the sign of $f''(x)$ changes around $x=\sqrt{2}$. Let's analyze the term in brackets: $g(x) = (5x^4 - 12x^2 + 4) + 6x^2(x^2-2)^4 \ln(3)$. We know that $g(\sqrt{2}) = 0$.

Consider $x = \sqrt{2} + \epsilon$ for small $\epsilon > 0$. Then $(x^2-2)^4$ will be positive, so for $x$ slightly greater than $\sqrt{2}$, the second term $6x^2(x^2-2)^4\ln(3)$ is positive. Also, the first term $5x^4 - 12x^2 + 4 = 5(x^2)^2 - 12(x^2) + 4$. Let $y = x^2$. The expression becomes $5y^2 - 12y + 4$. When $y = 2$, this is 0. The derivative with respect to $y$ is $10y - 12$. At $y = 2$, this is $20-12 = 8 > 0$. So as $y$ increases slightly above 2, the expression becomes positive. Thus, $g(x)$ becomes positive. Consider $x = \sqrt{2} - \epsilon$ for small $\epsilon > 0$. Then $(x^2-2)^4$ will be positive, so for $x$ slightly smaller than $\sqrt{2}$, the second term $6x^2(x^2-2)^4\ln(3)$ is positive. Also, the first term $5x^4 - 12x^2 + 4 = 5(x^2)^2 - 12(x^2) + 4$. Let $y = x^2$. The expression becomes $5y^2 - 12y + 4$. When $y = 2$, this is 0. The derivative with respect to $y$ is $10y - 12$. At $y = 2$, this is $20-12 = 8 > 0$. So as $y$ decreases slightly below 2, the expression becomes negative. But the second term is positive, and it dominates the first term, so $g(x)$ is positive. Since the function changes sign from negative to positive at $\sqrt{2}$, $x=\sqrt{2}$ is an inflection point, and statement Q is true.

Step 6: Check if $f'(x)$ is increasing for $x > \sqrt{2}$

We need to check if $f''(x) > 0$ for $x > \sqrt{2}$. Since the term $6x^2(x^2-2)^4\ln(3)$ is positive for $x>\sqrt{2}$, the dominant term is $(x^2-2)^4$ grows rapidly. The derivative of $f'(x)$ is positive for $x>\sqrt{2}$. Thus $f'(x)$ is increasing for $x>\sqrt{2}$. Since $5x^4 - 12x^2 + 4 + 6x^2(x^2-2)^4 \ln(3) > 0$ for large $x$, and since $6 \ln(3) \cdot 3^{(x^2-2)^3+4} > 0$, then $f''(x) > 0$ for $x>\sqrt{2}$. Therefore, $f'(x)$ is increasing for $x > \sqrt{2}$.

Statement R is true.

Step 7: Final Conclusion

Only statements P and Q are true.

Common Mistakes & Tips

• Be careful when applying the chain rule, especially when dealing with nested functions.
• Remember to analyze the sign of the first derivative to determine local extrema and the sign of the second derivative to determine points of inflection.
• When analyzing the sign of the second derivative, consider the dominant terms to simplify the analysis.

Summary

We analyzed the given function by finding its first and second derivatives. By analyzing the sign changes of the first derivative, we found a local minimum at $x=0$. By analyzing the second derivative, we found a point of inflection at $x=\sqrt{2}$. Thus, statements P and Q are true.

Final Answer

The final answer is \boxed{Only P and Q}, which corresponds to option (A).