If $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$ Then sum of all digits of $f(6)$ 
$(1):$ If $P(x)$ is a polynomial of Degree $4$ such that $P(-1) = P(1) = 5$ and
$P(-2)=P(0)=P(2)=2\;,$Then Max. value of $P(x).$
$(2):$ If $f(x)$ is a polynomial of degree $6$ with leading Coefficient $2009.$ Suppose
further that $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$
Then sum of all digits of $f(6)$ is

$\bf{My\; Try\; For\; (1):}$ Given $x=-2\;,x=0\;,x=+2$ are the roots of $P(x)=0.$
So $(x+2)\;,(x-0)\;,(x-2)$ are factors of $P(x)=0$. So we can write $P(x)$ as
$P(x) = A\cdot x\cdot (x-2)\cdot (x+2)(x-r)\;,$ So we have calculate value of $A$ and $r$
Now put $x=-1\;,$ we get $P(-1)=-3A\cdot (1+r)\Rightarrow -3A\cdot (1+r)=5............................(1)$
Now put $x=1\;,$ we get $P(1)=-3A\cdot (1-r)\Rightarrow -3A\cdot (1-r)=5..................................(2)$
So from $(1)$ and $(2)\;,$ We get $r=0$ and $\displaystyle A=-\frac{5}{3}.$
So Polynomial $\boxed{\boxed{\displaystyle P(x)=-\frac{5}{3}\cdot x^2\cdot (x^2-4)}}$
$\bf{My\; Try\; For \; (2):}$Let $f(x)=2x-1\;\forall\;  x=1\;,2\;,3\;,4\;,5.$
So we can say that $(x-1)\;,(x-2)\;,(x-3)\;,(x-4)\;,(x-5)$ are the roots of $f(x)-2x+1=0$
So $f(x)-2x+1=2009\cdot \underbrace{(x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)}\cdot\underbrace{(x-r)}$
Now How can i solve after that
Help me and plz explain me, is my $(1)$ Try is right or not
Thanks
 A: Question 1 directly states that $P(-2)=P(0)=P(2)=2$. Why on earth are you then saying that they're roots of the polynomial?
What you can say is that $x=-2,0,2$ are roots of $P(x)=2$, or equivalently $P(x)-2=0$. So we can say that, if $P(x)$ is 4th degree, then $P(x)-2$ is as well, and hence $P(x)-2=A(x-\alpha)(x-2)x(x+2)$ for some $A,\alpha$. Then you can insert your values for $P(1)$ and $P(-1)$.
Also, you haven't answered the question. Read it carefully, see what it's asking, and make sure you answer it.
As for 2, you're on the right track. The next step would probably be to look at what the derivative of that expression will look like at $x=2$.
A: For part 2, here's one way to simplify the differentiation. We have
$$f(x)-2x+1=2009\cdot (x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)\cdot(x-r)$$
Substituting $x + 3 \to x$,
$$f(x + 3) - 2(x+3)+1 = 2009(x+2)(x+1)(x)(x-1)(x-2)(x+3-r)$$
$$f(x + 3) -2x - 5 = 2009x(x^2-1)(x^2-4)(x + 3 -r)$$
$$f(x + 3) - 2x - 5 = 2009(x^5 - 5x^3 + 4x)(x + 3 -r)$$
Applying the chain rule to LHS and product rule to RHS,
$$f'(x+3) - 2 = 2009(5x^4 - 15x^2 + 4)(x + 3 - r) + 2009(x^5 - 5x^3 + 4x)$$
Substituting $x = -1$,
$$f'(2) - 2 = 2009(5 - 15 + 4)(2-r) + 2009(1-5+4)$$
$$0 = 2009\cdot(-6)\cdot(2-r)$$
to give $r=2$. Hence,
$$f(6) - 12 + 1 = 2009\cdot5!\cdot4$$
$$f(6) = 964331$$
