Mathematics - Application of Derivative Question with Solution | TestHub

Since $f(x)$ has local maxima at $x=-1$ and $f^{\prime }(x)$ has local minima at $x=0$. $$\begin{array}{rl}f\wedge (x)& =\lambda x\\ f^{\prime }(x)& =\lambda \frac{x^2}{2}+c\\ \frac{\lambda }{2}+c& =0\Rightarrow \lambda =-2c\end{array}$$ $$\left[f^{\prime }(-1)=0\right]$$ Again, On integrating both sides, we get $$\begin{array}{rl}& f(x)=\lambda \frac{x^3}{6}+cx+d\\ & f(2)=\lambda \left(\frac{8}{6}\right)+2c+d=18\\ & f(1)=\frac{\lambda }{6}+c+d=-1\end{array}$$ and $\therefore$ Using Eqs. (i), (ii) and (iii), we get $$\begin{array}{rl}f(x)& =\frac{1}{4}\left(19x^3-57x+34\right)\\ \therefore f^{\prime }(x)& =\frac{1}{4}\left(57x^2-57\right)\\ & =\frac{57}{4}(x-1)(x+1),\text{ using number line rule }\end{array}$$ $\therefore f(x)$ is increasing for $[1,2\sqrt{5}]$ and $f(x)$ is increasing for $[1,2\sqrt{5}]$ and $f(x)$ has local maximum at $x=-1$ and $f(x)$ has local minimum at $Cx=1$. Also, $$f(0)=\frac{34}{4}\text{. }$$