We are given that $f$ be a polynomial of degree $n \geq 3$ for which all of its roots are non-positive reals i.e. roots lies in the interval $(-\infty,0]$ and it is also given that $f(1)=1$ and we need to find the maximum possible value of $f’(1).$
We start with $n=3$ and then go for $n=4$ and the idea is same for all $n \geq 3.$ And once we understand it for $n=3,$ we can easily extend it to $n > 3.$
Suppose, roots of the polynomial with degree $3$ i.e. a cubic polynomial are $\alpha’, \beta’,\gamma’$ and so we can write the polynomial as:
$f(x)=(x-\alpha’)(x-\beta’)(x-\gamma’)$
Since, roots are non-positives which means $\alpha’ \leq 0,\beta’ \leq 0, \gamma’\leq0.$
Now, say $\alpha’=-\alpha, \beta’=-\beta,\gamma’=-\gamma$ which means $\alpha \geq 0, \beta\geq0,\gamma\geq0$
So, $f(x)$ is now transformed into $f(x)=(x+\alpha)(x+\beta)(x+\gamma)$ with $\alpha \geq 0, \beta\geq0,\gamma\geq0$.
And here, $f(1)=(1+\alpha)(1+\beta)(1+\gamma)$ and since, it is given that $f(1)=1$ so we need to divide $f(x)$ by $(1+\alpha)(1+\beta)(1+\gamma)$
Hence our final $f(x)$ is:
$$f(x)=\frac{(x+\alpha)(x+\beta)(x+\gamma)}{(1+\alpha)(1+\beta)(1+\gamma)}$$
Now $(x+\alpha)(x+\beta)(x+\gamma)$ is similar to $(x-\alpha)(x-\beta)(x-\gamma)$ which can be written as
$x^3 -(\alpha+\beta+\gamma)x^2 + (\alpha \beta + \beta \gamma + \gamma \alpha)x -\alpha \beta \gamma$ but here we using $+$ instead of $-$ so all terms will be same with $+$ sign.
So,
$$f(x)=\frac{x^3 + (\alpha+\beta+\gamma)x^2 + (\alpha \beta + \beta \gamma + \gamma \alpha)x +\alpha \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma)}$$
So $f’(x)$ is:
$$f’(x)=\frac{3x^2 + 2(\alpha+\beta+\gamma)x + (\alpha \beta + \beta \gamma + \gamma \alpha)}{(1+\alpha)(1+\beta)(1+\gamma)}$$
So,
$$f’(1)=\frac{3 + 2(\alpha+\beta+\gamma) + (\alpha \beta + \beta \gamma + \gamma \alpha)}{(1+\alpha)(1+\beta)(1+\gamma)}$$
Since we have to maximize $f’(1)$ it means we have to minimize the denominator of $f’(1)$ with positive sign and since $\alpha,\beta,\gamma \geq 0,$ So, $(1+\alpha)(1+\beta)(1+\gamma)$ will be minimum when $\alpha=\beta=\gamma=0.$
So, put $\alpha=\beta=\gamma=0$ in the expression of $f’(1)$ we get the maximum possible value os $f’(1)$ as $3$ which is the degree of the polynomial.
Similarly, for degree $4$ polynomial, we write:
$$f(x)=\frac{x^4 + (\alpha+\beta+\gamma+\delta)x^3 + (\alpha \beta + \beta \gamma + \gamma \delta + \delta \alpha + \alpha \gamma + \beta \delta)x^2 +(\alpha \beta \gamma + \beta \gamma \delta + \gamma \delta \alpha + \alpha \beta \delta)x+ \alpha \beta \gamma\delta}{(1+\alpha)(1+\beta)(1+\gamma)(1+\delta)}$$
Now, find $f’(x)$ using the similar procedure described above and we get the maximum possible value of $f'(1)$ as $4.$
We can extend it to n- degree polynomial and on the same idea since, derivative of $x^n$ is $nx^{n-1}$ and this multiplier $n$ decides the maximum possible value of $f’(1).$