Mathematics - Application of Derivative Question with Solution | TestHub

Since, $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$ and $p(x)$ is a real polynomial of least degree. Hence, let $p^{\prime }(x)=k(x-1)(x-3)=k\left(x^2-4x+3\right)$ $\Rightarrow p(x)=k\left(\frac{x^3}{3}-2x^2+3x\right)+c$ Now, $p(1)=6$ and $p(3)=2$ $\begin{array}{l}\Rightarrow \frac{4}{3}k+C=6\text{ and }0+C=2\Rightarrow k=3\\ \therefore p^{\prime }(x)=3(x-1)(x-3)\Rightarrow p^{\prime }(0)=9\end{array}$