calculus ivt

Question

Find the value of c that is guaranteed by the IVT, so that f (c)=k

 

F (x)=x^2+6/x^2+1  [0,2] k=5

Answer 1

The problem has 2 incorrect statements. The IVT (Intermediate value theorem) requires:

F(x) defined on [a,b] However [0,2] is incorrect, as F(0) is not defined. We will chose [1,2]

F(c)=k=5 is also incorrect. F(x) is always >5, so will select a different k.

F(1)=8; F(2)=6.5 From IVT, there is a c ∈ (a,b) so F(c)=k where k ∈ [F(a), F(b)]; select k=7 ∈ [6.5 , 8].

Calculating c: F(c) = c² +6/c² + 1 = 7; notation: c² = z; z+6/z-6=0; z²-6z+6=0, z₁=4; z₂=2;

One value of c=√2 =1.414 ∈ (1,2), as guaranteed per IVT.