3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

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

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 $P'(0) = 3A = 9\Rightarrow P'(0)=9$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.