A real valued function y = f(x) satisfies the | vedclass

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(B) Given the inequality: $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$. Let $g(x)

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$\frac{9}{2}$

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(B) Given the inequality: $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$. Let $g(x) = \frac{9}{4}x^2 + \frac{8}{9}$. We can rewrite the inequality as $f\left( x - \frac{4}{9} \right) \le g(x) - 2x$ and $g(x) + 2x \le f\left( x + \frac{4}{9} \right)$. Let $x - \frac{4}{9} = t$,then $x = t + \frac{4}{9}$. Substituting this,we get $f(t) \le \frac{9}{4}(t + \frac{4}{9})^2 + \frac{8}{9} - 2(t + \frac{4}{9}) = \frac{9}{4}t^2 + 2t + \frac{4}{9} + \frac{8}{9} - 2t - \frac{8}{9} = \frac{9}{4}t^2 + \frac{4}{9}$. Similarly,for the second part,let $x + \frac{4}{9} = t$,then $x = t - \frac{4}{9}$. Substituting this,we get $f(t) \ge \frac{9}{4}(t - \frac{4}{9})^2 + \frac{8}{9} + 2(t - \frac{4}{9}) = \frac{9}{4}t^2 - 2t + \frac{4}{9} + \frac{8}{9} + 2t - \frac{8}{9} = \frac{9}{4}t^2 + \frac{4}{9}$. Since $f(t) \le \frac{9}{4}t^2 + \frac{4}{9}$ and $f(t) \ge \frac{9}{4}t^2 + \frac{4}{9}$,it must be that $f(x) = \frac{9}{4}x^2 + \frac{4}{9}$. Now,find the second derivative: $f'(x) = \frac{d}{dx}\left( \frac{9}{4}x^2 + \frac{4}{9} \right) = \frac{9}{2}x$. $f''(x) = \frac{d}{dx}\left( \frac{9}{2}x \right) = \frac{9}{2}$. Therefore,$f''(2) = \frac{9}{2}$.

$A$ real valued function $y = f(x)$ satisfies the relation $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$. The value of $f''(2)$ is

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