Considering the real function

We know that the quadratic function is opened downwards because the term with degree 2 has negative coeffiient. Now to find the vertex of the function we can use the formula for axis of symmetry. -b/2a plugging in a=-2, b=4 we get 1. Now we just have to plug in x=1 to get the y coordinate of the vertex y=14.

-2 (x - 1)^2 < 0

-2 (x - 1)^2 +14 <14

F(x)<14

F(1) = 14

Therefore F admits a maximum in 1 , its value is 14

As the value of $\frac{d^2}{dx^2} f(x) < 0$ , therefore the maximum value of $f(x)$ exists at $\frac{d}{dx} f(x)=0$ .

$\frac{d}{dx} f(x)= -4x+4=0\Rightarrow x=1$

The maximum value is $f(1)=14$ .

$-2x^2+4x+12=14-(\sqrt{2}x-\sqrt{2})^2 \leq 14$ with equality when $x=1$