The local minimum value of the function \(f(x) = |x| + 2\), where x is a real number, is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
The function |x| + 2 is not differentiable at x = 0, so the derivative test cannot be used there. A direct comparison of values is enough.
Since the modulus is never negative:
\(|x| \ge 0\)
\(|x| + 2 \ge 2\)
and the equality holds only at x = 0, where
\(f(0) = 0 + 2 = 2\)
For every x near 0 (other than 0 itself), f(x) > 2 = f(0). So x = 0 is a point of local minimum, and the local minimum value is f(0) = 2.
Tip: The question asks for the value f(0) = 2, not the point x = 0 where it occurs; 0 is Option (1).