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What is the minimum value of y = \(\frac{ { x }^{ 2 } }{2}-3x\)? |
$9/2$ $-9/2$ $9/4$ $-9/4$ |
$-9/2$ |
The correct answer is Option (2) → $-9/2$ $f(x)=\frac{x^2}{2}-3x$ for extreme values, $f'(c)=0$ $⇒2x-3=0$ $⇒x=3/2$ $∴f(\frac{3}{2})=\frac{9}{8}-\frac{9}{2}$ $=-\frac{9}{2}$ |
