Minima?

f(x) = $2x^2 - x + 1$
f'(x) = $4x - 1$
Equate to 0:-
$4x - 1 = 0$
$x = \dfrac{1}{4}$

Substitute value of x in f(x)
= $2 × {\dfrac{1}{4}}^2 - \dfrac{1}{4} + 1$
= $2 × \dfrac{1}{16} + 1 - \dfrac{1}{4}$
= $\dfrac{7}{8} = \boxed{0.875}$

Lets apply second derivative test to show that the value of x obtained is minimum.

Then, we get the result $4$ which is positive, thus proving that the answer we got is the minimum value.

Ashish here it is, this function is similar to the function of your question. CD shows a negative slope(a tangent on the function curve is slope of the curve) while AB shows positive slope. Which leads the point in between them to be a minima.

In a similar way we can do for Maxima. Like this:

The formula of minimum is -D/4a

So the answer is 0.875

For the sake of variety, here's a third method which is applicable for quadratic functions only.

Given a quadratic function in the form of $f(x) = ax^2 + bx + c$ , the coordinates of the turning point is given as:

$\left(-\dfrac {b}{2a}, f \left( - \dfrac{b}{2a} \right)\right)$

To determine the type of turning point, just check the value of $a$ :

If $a>0$ , the turning point is a minimum point

If $a<0$ , the turning point is a maximum point

For the given function here, $a=2>0$ , therefore the turning point of the function is minimum. We then find the $x$ -coordinate of the minimum point:

$- \dfrac{b}{2a} = - \dfrac{-1}{2(2)} = \dfrac{1}{4}$

The minimum value is then given as:

$f \left(-\dfrac {b}{2a} \right) = f \left( \dfrac{1}{4} \right) = 2\left( \dfrac{1}{4} \right)^2 - \dfrac{1}{4} + 1 = \dfrac{7}{8} = \boxed{0.875}$

I find this method to be easier than completing the square, however, the best method is to use derivatives as shown in the solution above. The derivative method works not only for quadratic functions, but also for any other type of function. You just need to do an extra step to determine the type of turning point found.