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