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Let $f$ be a polynomial of degree $n \geq 3$ all of whose roots are non-positive real numbers. Suppose that $f(1)=1$. What is the maximum possible value of $f^{\prime}(1)?$ 

  1. $1$
  2. $n$
  3. $n+1$
  4. $\frac{n(n+1)}{2}$
  5. $f^{\prime}(1)$ can be arbitrarily large given only the constraints in the question
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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 $f’(1).$
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Given non positive root and maximum value of f'(1) 

means root should be zero (maximum value in non positive)

 consider degree 3 polynomial and roots : a,b,c

F(x) = X+ (a+b+c) x2 + (ab+bc+ca)x + abc

F'(x) = 3x2+ roots included values

for maximum a=b=c=0

F'(1)=3

now consider 4 degree polynomial

f(x) = x4+roots included values

for maximum roots =0

so f'(1)=4

.

.

similearly for n degree polynomial max f'(x)=n

Answer:

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