what is the min value of y = x^2 + x + 1 (The answer is y = 3/4)

Can someone explain the calculus approach to this question?

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

Step 1: Find critical point and for doing that take derivate of the function.

f **’** (x)= x^2 + x + 1 =0

so, we get : 2x +1 = 0 , x = -1/2

Now to determine y evaluate f at that critical value f (-1/2):

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

= 3/4

Now , we have to take double-derivative to check to check minimum and maximum

f **‘’** (x) = 2

Now , because this value is positive it tell us that f evaluated at critical value of -1/2 is the minimum.

so, our ans is 3/4

**Rules:**

The function has a minimum value at *x* = *a* if *f ’* ( *a* ) = 0

and *f ‘’* ( *a* ) = a positive number.

The function has a maximum value at *x* = *a* if *f ’* ( *a* ) = 0

and *f ‘’* ( *a* ) = a negative number.

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Thanks that was very helpful!