Intermediate Value Theorem

[Alexander Lorien]

Explain why

has a solution in the interval

using Intermediate Value Theorem.

 

[Harry_the_cat]

I would consider \(\displaystyle x^{10}-10x^2+5=0\).
Evaluate LHS when x=0 and x=1.

 

[Alexander Lorien]

I would consider \(\displaystyle x^{10}-10x^2+5=0\).
Evaluate LHS when x=0 and x=1.

This is reasonable. Since f(0) = 5 > 0. and f(1) = -4 < 0. So by IVT, if a = 0, b = 1, N=0, there exists c ∈ ( 0,1) such that f(c) = 0. HOWEVER, if i am going to consider the x=2 instead of 1, it differs. since f(2) = 989 > 0.

 

[Harry_the_cat]

Yes that's correct. But isn't it true that "a solution in the interval (0, 1)" means that there must be a solution in the interval (0, 2). So considering x=2 isn't useful here.

 

[Alexander Lorien]

Yes that's correct. But isn't it true that "a solution in the interval (0, 1)" means that there must be a solution in the interval (0, 2). So considering x=2 isn't useful here.

I see. Also I tried using the interval (1,2) and there is a solution So there are two solutions in the interval (0,2)

 

[Harry_the_cat]

I see. Also I tried using the interval (1,2) and there is a solution So there are two solutions in the interval (0,2)

At least 2?