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Find its maximum or minimum value. \[f(x)=\frac{ -x^2 }{ 3 }+2x+7\]
\[f'(x)=\frac{ -2x }{3 }+2\] f'(x)=0 gives \[\frac{ -2x }{ 3 }+2=0,\frac{ -2x }{ 3 }=-2,x=-2*\frac{ 3 }{ -2 }=3\] \[f''(x)=\frac{ -2 }{ 3 }\] At x=3, \[f''(x)=-\frac{ 2 }{ 3 }<0\] Hence there is maxima at x=3 there is no minimum value.
put x=3 in f(x) and get the maximum value.
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