<s>[INST] If $P(x)$ is a poly. of least degree and local Max. at $x=1$ and Local Min. at $x=3,$ Then $P'(0)$ is If $P(x)$ is a polynomial of least degree which has a local Maxima at $x=1$ and Local Minima at
$x=3.$ If $P(1)=6$ and $P(3)=2$. Then $P'(0)=$
$\bf{My\; Try::}$ Given function has one Maxima and one Minima So $P(x)$ must have least 
degree $3$ polynomial. So Let $P(x)=Ax^3+Bx^2+Cx+D$ and $P'(x)=3Ax^2+2Bx+C.$
Now Given $P(1)=6\Rightarrow A+B+C+D = 6....................(1)$
and Given $P(3)=2\Rightarrow 27A+9B+3C+D=2..............(2)$
and Given $P'(1)=0\Rightarrow 3A+2B+C = 0.........................(3)$
and Given $P'(3)=0\Rightarrow 9A+6B+C = 0.........................(4)$
Now Subtract $(4)-(3)$, We Get $6A+4B=0\Rightarrow 3A+2B=0$
Similarly Sub $(2)-(1)\;,$ We Get $26A+8B+2C=-4\Rightarrow 13A+4B+C=-2$
Is there is any other method my which we can solve the above question in less complex way.
plz explain me, Thanks
 [/INST] Continuing from what you have got,
Subtracting $13A + 4B + C = -2$ from $3A + 2B + C = 0$ yields $10A + 2B = -2$.
Now we have $3A + 2B = 0$ and $10A + 2B = -2$.
Hence $A = -2/7$ and $B$, $C$, $D$ can be calculated by putting back into $3A + 2B = 0$, $3A + 2B + C = 0$ and $A + B + C + D = 6$ in order.
Then $P'(0)$ can be easily obtained.
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