If S_1 and S_2 are respectively the sets of l | vedclass

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(A) Given the function $f(x) = 9x^4 + 12x^3 - 36x^2 + 25$.To find the critical points,we find the first derivative $f'(x)$:$f'(x) = 36x^3 + 36x^2 - 72

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$S_1 = \{-2, 1\}; S_2 = \{0\}$

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(A) Given the function $f(x) = 9x^4 + 12x^3 - 36x^2 + 25$.To find the critical points,we find the first derivative $f'(x)$:$f'(x) = 36x^3 + 36x^2 - 72x$Setting $f'(x) = 0$:$36x(x^2 + x - 2) = 0$$36x(x - 1)(x + 2) = 0$The critical points are $x = -2, 0, 1$.Now,we use the second derivative test $f''(x) = 108x^2 + 72x - 72$:For $x = -2$: $f''(-2) = 108(4) + 72(-2) - 72 = 432 - 144 - 72 = 216 > 0$ (Local Minima).For $x = 0$: $f''(0) = -72 For $x = 1$: $f''(1) = 108 + 72 - 72 = 108 > 0$ (Local Minima).Thus,the set of local minimum points $S_1 = \{-2, 1\}$ and the set of local maximum points $S_2 = \{0\}$.

If $S_1$ and $S_2$ are respectively the sets of local minimum and local maximum points of the function $f(x) = 9x^4 + 12x^3 - 36x^2 + 25, x \in R$,then

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