Application of DerivativeHard
Question
Least value of the function, f(x) = 2x2 - 1 + 
is ;
is ;Options
A.0
B.3/2
C.2/3
D.1
Solution
f′(x) = 
2x2 ≥ 1
2x2 + 1 - √2 ≥ 2 - √2 > 0
At x = 0, f(x) is least.
Least value = f(0) = 1
2x2 ≥ 1
2x2 + 1 - √2 ≥ 2 - √2 > 0
At x = 0, f(x) is least.
Least value = f(0) = 1
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