Math Problems
#Problems #Quest76
#Solutions | #Quest76
deg(P(x))=1 => P(x)=kx+c
For all x€R P(x) increasing.
Therefore max(P(x))=kb+c
min(P(x))=ka+c
maxP(x)-minP(x)=k(b-a) (1)
we have
maxP(x)-minP(x)=b-a (2)
From (1) and (2)
k(b-a)=b-a => k=1
P(x)=kx+c=x+c
c€R.
Answer: x+c , c€R
Author:Jamshidbek Hasanov
deg(P(x))=1 => P(x)=kx+c
For all x€R P(x) increasing.
Therefore max(P(x))=kb+c
min(P(x))=ka+c
maxP(x)-minP(x)=k(b-a) (1)
we have
maxP(x)-minP(x)=b-a (2)
From (1) and (2)
k(b-a)=b-a => k=1
P(x)=kx+c=x+c
c€R.
Answer: x+c , c€R
Author: