# 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

• If you have $P(-2) = P(0) = P(2) = 2 \neq 0$, why can you say that $x(x+2)(x-2)$ is a factor of $P(x)$? – Platehead Nov 14 '14 at 2:57

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 $P(-1)$.

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 $x=2$.

• Oh Sorry I did not Noticed that, – juantheron Nov 14 '14 at 4:08

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,

$$f(6) - 12 + 1 = 2009\cdot5!\cdot4$$ $$f(6) = 964331$$