Question

show that the equation x^3-3x^2+1=0has one root between 1and 2​

Answer 1

Answer:

Let f(x)=x3−3x+b

Let's suppose −1≤x<y≤1

Then f(y)−f(x)=y3−3y+b−(x3−3x+b)=(y−x)(x2+xy+y2−3)

The first factor y−x is positive. We have x2≤1,xy<1,y2≤1 so that x2+xy+y2−3 is negative.

This means that f(y)<f(x)

Suppose −1≤y,z≤1 and f(z)=0. If y<z then f(y)>f(z)=0 and if y>z then f(y)<f(z)=0

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Answer 2

Answer:

hope it's helpful.,..........