question_answer

If \[f(x)={{x}^{3}}+a{{x}^{2}}+bx+c\] has maximum value at \[x=-1\] and minimum at x = 3. Determine the values of a, b and c.

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\[c\in R\]