Application of DerivativeHard
Question
If f(x)
, for every nu8mber x, then the minimum value of f
, for every nu8mber x, then the minimum value of fOptions
A.does not exist because f is unbounded.
B.is not attained even though f is bounded
C.is equal to 1
D.is equal to - 1
Solution
Given, f(x)
f(x) will be minimum when
is maximum,
ie, when x2 + 1 is minimum
ie, at x = 0
∴ Minimum value of f(x) is f(0) = - 1
f(x) will be minimum when
is maximum, ie, when x2 + 1 is minimum
ie, at x = 0
∴ Minimum value of f(x) is f(0) = - 1
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