Let $f(x)=3^{(x^2-2)^3+4}.$

We must test the statements:

• $P$: $x=0$ is a point of local minima of $f$
• $Q$: $x=\sqrt2$ is a point of inflection of $f$
• $R$: $f'$ is increasing for $x>\sqrt2$

1. Rewrite the function

Let $g(x)=(x^2-2)^3+4.$ Then $f(x)=3^{g(x)}.$

Since $3^u$ is strictly increasing in $u$, many features of $f$ can be studied using $g$.

2. First derivative

Using $\frac{d}{dx}(3^{g(x)})=3^{g(x)}\ln 3\, g'(x),$ we first compute $g'(x)$: $g'(x)=3(x^2-2)^2\cdot 2x=6x(x^2-2)^2.$ Hence $f'(x)=3^{(x^2-2)^3+4}(\ln 3)\cdot 6x(x^2-2)^2.$ So $f'(x)=6(\ln 3)\,x(x^2-2)^2\,3^{(x^2-2)^3+4}.$

Because $3^{(x^2-2)^3+4}>0$, $(x^2-2)^2\ge 0$, and $\ln 3>0$, the sign of $f'(x)$ is determined by $x$.

Thus:

• for $x<0$, $f'(x)<0$ (except at $x=-\sqrt2$ where $f'=0$),
• for $x>0$, $f'(x)>0$ (except at $x=\sqrt2$ where $f'=0$).

So the function decreases for $x<0$ and increases for $x>0$.

Therefore, $x=0$ is a local minimum point.

So $P$ is true.

3. Check inflection at $x=\sqrt2$

We need $f''(x)$.

Write $f'(x)=3^{g(x)}\ln 3\, g'(x).$ Then $f''(x)=3^{g(x)}\left[(\ln 3)^2(g'(x))^2+(\ln 3)g''(x)\right].$ So $f''(x)=3^{g(x)}\ln 3\left[(\ln 3)(g'(x))^2+g''(x)\right].$

Now compute $g''(x)$: $g'(x)=6x(x^2-2)^2.$ Differentiate: $g''(x)=6(x^2-2)^2+6x\cdot 2(x^2-2)\cdot 2x$ $=6(x^2-2)^2+24x^2(x^2-2)$ $=6(x^2-2)\big[(x^2-2)+4x^2\big]$ $=6(x^2-2)(5x^2-2).$

Hence $f''(x)=3^{g(x)}\ln 3\left[(\ln 3)(6x(x^2-2)^2)^2+6(x^2-2)(5x^2-2)\right].$

Now examine near $x=\sqrt2$. At $x=\sqrt2$, we have $x^2-2=0$, so $g'(\sqrt2)=0,\qquad g''(\sqrt2)=0,$ and therefore $f''(\sqrt2)=0.$

But to test inflection, we need sign change of $f''$ around $x=\sqrt2$.

Near $x=\sqrt2$, the term $(\ln 3)(g'(x))^2=(\ln 3)\cdot 36x^2(x^2-2)^4$ is of order $(x^2-2)^4$, whereas $g''(x)=6(x^2-2)(5x^2-2)$ is of order $(x^2-2)$. So near $x=\sqrt2$, the sign of the bracket is governed by $g''(x)=6(x^2-2)(5x^2-2).$ Since near $x=\sqrt2$, $5x^2-2>0$, the sign of $f''(x)$ is essentially the sign of $(x^2-2)$.

Thus:

• for $x<\sqrt2$ and close to $\sqrt2$, $x^2-2<0 \implies f''(x)<0$,
• for $x>\sqrt2$ and close to $\sqrt2$, $x^2-2>0 \implies f''(x)>0$.

So $f''$ changes sign at $x=\sqrt2$.

Therefore, $x=\sqrt2$ is a point of inflection.

So $Q$ is true.

4. Check whether $f'$ is increasing for $x>\sqrt2$

For $f'$ to be increasing, we need $f''(x)>0 \quad \text{for all } x>\sqrt2.$

From above, $f''(x)=3^{g(x)}\ln 3\left[(\ln 3)(g'(x))^2+g''(x)\right].$ For $x>\sqrt2$:

• $3^{g(x)}>0$,
• $\ln 3>0$,
• $(\ln 3)(g'(x))^2\ge 0$,
• and $g''(x)=6(x^2-2)(5x^2-2).$

If $x>\sqrt2$, then $x^2-2>0$ and also $5x^2-2>5\cdot 2-2=8>0.$ Hence $g''(x)>0.$ Therefore the whole bracket $(\ln 3)(g'(x))^2+g''(x)>0.$ So $f''(x)>0 \qquad \text{for all } x>\sqrt2.$ Thus $f'$ is increasing for $x>\sqrt2$.

So $R$ is true.

5. Final conclusion

All three statements $P, Q, R$ are true.

Therefore the correct option is: $\boxed{D\text{: All P, Q and R}}$

6. Comparison with stored answer

Stored correct answer: $D$

Our derived answer: $D$

So the derived answer agrees with the stored answer.