A) \[-\frac{1}{4}\]
B) \[\frac{3}{2}\]
C) \[\frac{-9}{8}\]
D) \[\frac{9}{4}\]
Correct Answer: C
Solution :
\[f(x)=2{{x}^{2}}+x-1\] Þ \[f'(x)=4x+1\Rightarrow f'(x)=0\Rightarrow x=-\frac{1}{4}\] \[{f}''\,(x)=4=+ve\] \[\therefore {{[f(-1/4)]}_{\min }}=\frac{2}{16}-\frac{1}{4}-1=\frac{-9}{8}\].You need to login to perform this action.
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