Question:
Construct a polynomial $g\left ( x \right )$ with the following properties:
$g\left ( 0 \right )=\frac{3}{2}, g'\left ( 0 \right )=-4$ and $g'\left ( x \right )=0$ at $x=-1,1,2,4$
Hint: $g'\left ( x \right )=\alpha \left ( x-\beta \right )\left ( x-y \right )\cdots \left ( x-\delta \right )$
Attempt:
$g'\left ( x \right )=0 \implies \int g'\left ( x \right ).dx=C$
But $g\left ( 0 \right )=\frac{3}{2} \implies C=\frac{3}{2}$
The general idea I have is to form a differential equation by adding together the above first derivative and function and then using a software package to evaluate.
I'm not sure how the hint help. But some hints from the Mathstack community would be helpful.
Thanks in advance.
The derivative function $\,g'(x)\,$ is a polynomial, and its zeros are given. Thus: $$ \begin{align} g'(x) &= \alpha\,(x+1)(x-1)(x-2)(x-4) \\[2mm] g'(0) &= \alpha\,(0+1)(0-1)(0-2)(0-4) =-4\quad\Rightarrow\,\alpha=\frac12 \\[2mm] g'(x) &= \frac{1}{2}\,(x+1)(x-1)(x-2)(x-4) \\[2mm] &= \frac12x^4-3x^3+\frac72x^2+3x-4 \end{align} $$ Hence: $$ g(x) =\int g'(x)\,dx = \color{red}{\frac1{10}x^5-\frac34x^4+\frac76x^3+\frac32x^2-4x+\frac32} $$ Where we use $\,g(0)=3/2\,$ to calculate the constant.