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The absolute maximum value of the function f(x)=x2+2xf(x) = \frac{x}{2} + \frac{2}{x} in the interval [−3,3][-3, 3], is

Solution

✅ Correct Option: 3

Setting f′(x)=12−2x2=0f'(x)=\frac{1}{2}-\frac{2}{x^2}=0 gives x=±2x=\pm 2, with f(2)=2f(2)=2 and f(−2)=−2f(-2)=-2. The endpoints give f(3)=136f(3)=\frac{13}{6} and f(−3)=−136f(-3)=-\frac{13}{6}. The largest of these values is 136\frac{13}{6}.

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