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