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


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.

  • $\begingroup$ To play safe I think you should also check $P''(1) < 0$ and $P''(3) > 0$. $\endgroup$ – Empiricist Nov 14 '14 at 4:36

Thanks Friends ,I have got it

My Solution:: Let $P'(x) = A\cdot (x-1)\cdot (x-3)\;,$ Then Integrate both side w. r. to $x\;,$ We Get

$\displaystyle P(x) = A\left[\frac{x^3}{3}-2x^2+3x\right]+\mathcal C\;,$ Now for calculation of $A$ and $\mathcal C\;,$ Put $x=1$ and $x=3.$

So $\displaystyle P(1) = \frac{4A}{3}+\mathcal C\Rightarrow 4A+3\mathcal C = 18$ and $\displaystyle P(3) = \mathcal C\Rightarrow \mathcal C = 2.$ Now Solve These

Equation we get $\displaystyle A=3$ and $\displaystyle \mathcal C = 2$. So $P'(0) = 3A = 9\Rightarrow P'(0)=9$


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