Answer:
We have, \[f\,(x)={{x}^{3}}+a{{x}^{2}}+bx+c\] On differentiating f (x)w.r.t. x, we get \[f'x=3{{x}^{2}}+2ax+b\] For maximum or minimum, put \[f'(x)=0\] \[f'(-\,1)=3\,{{(-\,1)}^{2}}+2a\,(-\,1)+b=0\] \[\Rightarrow \] \[3-2a+b=0\] \[\Rightarrow \] \[2a-b=3\] ?(i) For maximum or minimum, put \[f'(x)=0\] \[f'(3)=3\,{{(3)}^{2}}+2a\,(3)+b=0\] \[\Rightarrow \] \[27+6a+b=0\] \[\Rightarrow \] \[6a+b=-\,27\] ?(ii) On adding Eqs. (i) and (ii), we get \[\Rightarrow \] \[a=-\,3\] Put in\[a=-\,3\] Eq. (i), we get \[2\,(-\,3)-b=3\] \[\Rightarrow \] \[-\,6-b=3\] \[\Rightarrow \] \[-\,b=-\,3+6=9\] \[\therefore \] \[b=-\,9\]and \[c\in R\] Thus, \[a=-\,3,\]\[b=-\,9\]and\[c\in R\]
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