Alexander Lorien
New member
- Joined
- Oct 2, 2020
- Messages
- 9
Explain why
has a solution in the interval
using Intermediate Value Theorem.
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.I would consider \(\displaystyle x^{10}-10x^2+5=0\).
Evaluate LHS when x=0 and x=1.
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)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.
At least 2?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)