# What is the local maximum and or miminum of the function f(x)=2x^2+2x^2-4x?

Feb 15, 2015

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$f \left(x\right) = 2 {x}^{2} + 2 {x}^{2} - 4 x$
is equivalent to
$f \left(x\right) = 4 {x}^{2} - 4 x$

$f ' \left(x\right) = 8 x - 4$
At the local maximum/minimum
$f ' \left(x\right) = 0$
that is
$8 x - 4 = 0$ $\rightarrow$ x = 1/2#

$f \left(\frac{1}{2}\right) = - 1$
Since the coefficient of ${x}^{2}$ is positive
this is a local minimum
which occurs at $\left(x , f \left(x\right)\right) = \left(\frac{1}{2} , - 1\right)$