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!