Question

P(x) is polynomial of degree 3 which have maximum value 8 at x = 1, min 6 at x = 2 find P(0) ​

Answer 1

Given :

P(x) is polynomial of degree 3 which have maximum value 8 at x = 1, and min 6 at x = 2

To Find :

The value of p(0)

Solution :

Given : 2,1 are the critical points of p(x)

Thus,

$\sf\:p(x)'=\lambda(x-1)(x-2)$

Where ,λ is a constant

$\sf\:p(x)'=\lambda(x^2-2x-x+2)$

$\sf\:p(x)'=\lambda(x^2-3x+2)$

Now integrate

$\sf\:p(x)=\lambda\int(x^2-3x+2)$

$\sf\:p(x)=\lambda(\frac{x^3}{3}-\frac{-3x^2}{2}+2x)+c$

It is given that p(1)=8 then ,

$\sf\:p(1)=\lambda(\frac{1^3}{3}-\frac{-3\times1^2}{2}+2)+c$

$\sf\:p(1)=\lambda(\frac{1}{3}-\frac{-3}{2}+2)+c$

$\sf\:p(1)=\lambda(\frac{14-9}{6})+c$

$\sf\:p(1)=\frac{5\lambda}{6}+c$

$\sf\:8=\frac{5\lambda}{6}+c....(1)$

It is also given that p(2)=6 ,then

$\sf\:p(2)=\lambda(\frac{2^3}{3}-\frac{-3\times2^2}{2}+2\times2)+c$

$\sf\:p(2)=\lambda(\frac{8}{3}-\frac{-3\times4}{2}+4)+c$

$\sf\:p(2)=\lambda(\frac{8-6}{3})+c$

$\sf\:p(2)=\frac{2\lambda}{3}+c$

$\sf\:6=\frac{2\lambda}{3}+c....(2)$

Subtract Equation (2) from (1)

Then ,

$\sf\:2=\dfrac{5\lambda}{6}-\dfrac{2\lambda}{3}$

$\sf\:2=\dfrac{(5-4)\lambda}{6}$

$\sf\:2=\dfrac{\lambda}{6}$

$\sf\lambda=12$

Put ,λ=12 in equation (1)

$\sf\:8=\dfrac{5}{6}\times12+c$

$\sf\:8=10+c$

$\sf\:c=-2$

Thus,

$\sf\:p(x)=12(\frac{x^3}{3}-\frac{-3x^2}{2}+2x)-12$

We have to find p(0)

Then ,

$\sf\:p(0)=12(\frac{0^3}{3}-\frac{-3\times0^2}{2}+2\times0)-2$

$\sf\:p(0)=-2$

Therefore ,p(0)= -2

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Since P(x) is a polynomial of degree 3

Let

$P(x) = a {x}^{3} + b {x}^{2} + cx + d \: \: \: \: ......(1)$

Where a, b, c, d are constants

$P'(x) = 3a {x}^{2} + 2bx + c$

Now by the given condition 1 :

O(x) has a maximum value 8 at x= 1

$P(1) = 8$

$\implies \: a + b + c + d = 8 \: \: \: ....(2)$

And

$P'(1) = 0$

$\implies \: 3a + 2b + c = 0 \: \: \: ....(3)$

Again by the given condition 2 :

P(x) has a minimum value 6 at x = 2

$P(2) =6$

$\implies \: 8a + 4b + 2c + d = 6 \: \:...... (4)$

And

$P'(2) = 0$

$\implies \: 12a + 4b + c = 0 \: \: \: ....(5)$

Equation 4 - Equation 2 gives

$7a + 3b + c = - 2 \: \: \: ...(6)$

Equation (5) - Equation 3 gives

$9a + 2b = 0 \: \: .......(7)$

Equation 5 - Equation 6 gives

$5a + b \: = 2 \: \: \: ....(8)$

Solving Equation 7 & Equation 8

$a = 4 \: \: \: and \: \: b = - 18$

From Equation 5

$c = - 12a - 4b = - 48 + 72 = 24$

From Equation 2

$d = 8 - 4 + 18 - 24 = - 2$

So

$P(x) = 4 {x}^{3} - 18 {x}^{2} + 24x - 2 \: \: \:$

So

P(0) = 0 - 0 + 0 - 2 = - 2