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Application Of Derivatives — Easy JEE math MCQ

The minimum value of $f(x) = x^2 + 4x + 5$ is:
  1. A. $1$
  2. B. $0$
  3. C. $5$
  4. D. $4$

Solution

$f'(x) = 2x + 4 = 0 \Rightarrow x = -2$ $f(-2) = 4 - 8 + 5 = 1$

MATH

easy • PYQ Reworded
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The minimum value of f(x)=x2+4x+5f(x) = x^2 + 4x + 5 is:
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Application Of Derivatives — Easy JEE Mathematics MCQ | MyGoalPrep