To determine which of the statements P, Q, and R are true, we need to analyze the function \( f(x) = 3^{(x^2 - 2)^3 + 4} \) and evaluate each statement step-by-step.
Step 1: Analyze Statement P
Statement P claims that \( x = 0 \) is a point of local minima of \( f \).
To find the point of local minima, we should find the first derivative \( f'(x) \) and check the sign changes around \( x = 0 \).
The derivative of \( f(x) \) is given by:
f'(x) = 3^{g(x)} \ln(3) \cdot g'(x), where \( g(x) = (x^2 - 2)^3 + 4 \).
\begin{align*} g'(x) &= 3(x^2 - 2)^2 \cdot 2x = 6x(x^2 - 2)^2. \end{align*}
At \( x = 0 \), \( g'(0) = 6 \cdot 0 \cdot (-2)^2 = 0 \), but to find if it is a minimum, check the second derivative \( f''(x) \):
f''(x) \propto g''(x) = 6(2(x^2 - 2) \cdot 2x + (x^2 - 2)^2 \cdot 2) = 12(x^2 - 2)(x) + 12x(x^2 - 2)^2.
Since at \( x = 0 \), \( g''(x) = 0 \), \( f''(0) \gt 0 \) indicates it is indeed a local minimum.
Step 2: Analyze Statement Q
Statement Q claims that \( x = \sqrt{2} \) is a point of inflection of \( f \).
A point of inflection occurs where the second derivative changes sign, so we calculate \( f''(x) \) and see if there’s a sign change at \( x = \sqrt{2} \).
g'(\sqrt{2}) = 6\sqrt{2}(0) = 0 and find \( g''(\sqrt{2}) \).
Now consider the pattern \( g''(x) \) for \( x \approx \sqrt{2} \). There is indeed a change in concavity, confirming it is a point of inflection.
Step 3: Analyze Statement R
Statement R claims \( f' \) is increasing for \( x \gt \sqrt{2} \).
For this, we need to ensure that the third derivative \( f'''(x) \) is positive for \( x \gt \sqrt{2} \).
Given \( g'(x) = 6x(x^2 - 2)^2 \), for \( x \gt \sqrt{2} \), since \( x^2 \gt 2 \), we conclude \( g'(x) \) is always increasing.
Thus, all conditions in P, Q, and R are satisfied.
The correct option is All P, Q, and R.