Inspired by Rohit Kumar

$f(x)=x^2+(x+1)^2=2x^2+2x+1$ and its vertex is $-\frac{b}{2a}=-\frac{1}{2}$ . Therefore, the minimum is $f(-\frac{1}{2})$ ,

$\implies$ $\large{\boxed{\frac{1}{2}}}$ .

$x^{2} + (x+1)^{2} = 2x^{2}+2x+1$ $= 2\Big( x+\dfrac{1}{2}\Big)^{2} + \dfrac{1}{2}$ The minimum value is achieved when the term which is squared is $0$ . Hence the minimum value is $\dfrac{1}{2}$ .

Let $f$ be the function such that $f(x) = x^{2} + (x+1)^{2}$ . Then the first derivative is

$f'(x) = 4x + 2$

But $f'(x) = 0$ gives is an extreme point at $x = -1/2$ . The second derivative is

$f''(x) = 4 > 0$

But $f''(x) > 0$ means that the point we chose is a minimum. Then the minimum value of the function $f$ happens at $x = -1/2$ and

$f(-1/2) = \frac{1}{2}$

$f(x)=y=x^2+(x+1)^2=2x^2+2x+1$

$\frac{dy}{dx}=4x+2$

You obtain your max. or min. points if there are any at $\frac{dy}{dx}=0$

Then, $4x+2=0, \boxed{x=-\frac12}$

Now substitute by $x$ value in $f(x)$

You get $\boxed{y=\frac12}$

Note: this a minimum value because $f''(x)>0$ (a positive number)

Substitute $y=x+\dfrac{1}{2}$ to get $\left(y-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=2y^2+\dfrac{1}{2}$

But by the Trivial inequality, $2y^2+\dfrac{1}{2}\ge \dfrac{1}{2}$ so we are done.

The minimum is achieved when $x^{2} = (x+1)^{2}$ - this is the equality case for AM-GM. (It can also be proven that the minimum is achieved using a multitude of other ways, including possibly Cauchy-Schwarz and even geometrically.)

Since $x \neq x+1$ , it must be that $-x = x+1$ , ie $x = \boxed{-\frac{1}{2}}$ .

f(x) = x^2+ x^2+2x+1 = 2x^2+2x+1

Since coefficient of x^2 is positive, f(x) has a minimum turning point.

Turning point is at x= -2/2(2) = -1/2

So minimum is f(-1/2)

f(-1/2) = 1/2

So minimum value is 1/2

$x ^ { 2 } + ( x + 1 ) ^ { 2 }$

$y = 2x ^ { 2 } + 2x + 1$

$y = 2 ( x + \frac { 1 } { 2 } ) ^ { 2 } + \frac { 1 } { 2 }$

Hence, $\frac { 1 } { 2 }$ .

Sqrt(x^2+(x+1)^2) is the distance between (x,x) which is a point on line y=x and (0,-1). By graph the shortest distance is half the unit square's diagnol or 1/sqrt(2). Thus tye answer is 1/2

The function is $2x^2+2x+1$ and its minimum is $-\frac D {4a} = \frac 1 2$ .

To find a minimum or a maximum, we need the first derivative of the function, which is $4x+2$ , we set equal to 0 and solve for $x$ ; that is, $2x+2(x+1)=0$ , we obtain $x=(\frac{-1}{2}$ ), then $f(-1/2)=1/2$

x^2+(x+1)^2=2x^2+2x+1

a[x+b/(2a)]^2+(4ac-b^2)/4a

minimum y-value=(4ac-b^2)/4a

which is [4(2)(1)-4]/8=1/2

Obvoiusly: the function which quadratic function, let f(x)=2x^2 +2x +1, f will be minimum when x = -1/2. Hence, f_min= 1/2

Use calculus: take derivative of function; then set derivative equal to zero; finally, solve for x.

Easy solution. 2X^2 + 2X +1 = 2X^2 + 2 (squar root of 2)X 1/(square root of 2) + 1/2 + 1/2 = ((square root of 2)X + 1/(square root of 2))^2 + 1/2 >= 1/2

For symmetry reasons, it's gotta be in the middle between $-1$ and $0$ , and so we have ${ \left( -\frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 2 }$