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Find the minimum value of $3-4x+2x^2$.
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(&le; 720)

what is this?

$f(x) = 3-4x+2x^2$

$f '(x) = -4 + 4x = 0 \implies x=1$

$f ''(x) = 4$

$f ''(1) = 4>0$, therefore at $x=1$ we will get minimum value, which is : $3 - 4(1) + 2(1)^2 = 1.$
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