At what value of x is the function x\(^2\) + x + 1 minimum?
- -1
-
\(-\frac{1}{2}\)
- \(\frac{1}{2}\)
- 1
The correct answer is: B
Explanation
x\(^2\) + x + 1
\(\frac{dy}{dx}\) = 2x + 1
At the turning point, \(\frac{dy}{dx}\) = 0
2x + 1 = 0
x = -\(\frac{1}{2}\)