Question

In: Math

Maxima and Minima.

Let p(x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x = 3. If p(1) = 6 and p(3) = 2, then p’(0) is

Solutions

Expert Solution

Let p’(x) = k(x-1)(x-3)

= k(x² – 4x + 3)

p(x) = k(x³/3 – 2x² + 3x) + c (Integrating)

 

Given p(1) = 6.

=> k(⅓ – 2 + 3) + c = 6

=> 4k/3 + c = 6 …(i)

Also given p(3) = 2

=> k(9 – 18 + 9) + c = 2

=> c = 2

 

Put c in (i), we get

4k/3 + 2 = 6

=> 4k/3 = 4

=> k = 3

 

So p’(x) = 3(x-1)(x-3)

p’(0) = 3(0-1)(0-3)

= 9

 

 


Value of p’(0) is 9.

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