Question

Intermediate Value Theorem

Answer 1

I have 4 problems I am struggling with. My instructions are to determine if the IVT is applicable or not and state why or why not. If it is applicable, find the value of c guaranteed to exist by the theorem.

Answer 2

\[f(x)=\frac{ x-3 }{ x+2}\] on the interval [-1,2] for \[f(c)=\frac{ 2 }{ 3 }\]

Answer 3

Ok for this one I put that f is continuous on [-1,3]

Answer 4

@Hero @zepdrix @jim_thompson5910 anyone familiar with IVT

Answer 5

\[\frac{ x-3 }{ x+2 }=\frac{ 2 }{ 3 }\]
x=13

Answer 6

doesn't it fail since 13 is in the interval?

Answer 7

f(-1)=-4 and f(3)=0
doesn't f(13) need to be in interval [-1,3]

Answer 8

ok maybe if I state my second and third question it will help us with problem 1

Answer 9

\[f(x)=\frac{ x-3 }{ x+2}\]
on the interval [-4,1] for f(c)=2/3

Answer 10

Ok for this one I said IVT is not applicable since f(x) is not continuous from [-3,1] at x=-2.

Answer 11

problem 3
\[f(x)=\frac{ x }{ x-2}\] on the interval [-1,1] for f(c)=-1/2

Answer 12

My response is IVT is applicable since f(x) is continuous on [-1,1]
then I solved for c
\[\frac{ x }{ x-2 }=\frac{ -1 }{ 2 }\]
and x=2/3
and since f(-1)=1/3 and f(1)=-1
f(2/3)=-1/2 is in the interval [-1,1]

Answer 13

Oh oh here is the problem with the first one maybe.
\(\Large\rm f(-1)=-4\)
\(\Large\rm f(\quad2)=-1/4\)

The IVT guarantees that the function will take on any value between -4 and -1/4 between x=-1 and x=2.

It doesn't tell us anything about the function taking on the value 2/3 in this interval.
Am I thinking about that correctly? I hope?

Answer 14

sorry, I can't type today, I meant [-1,3] as the interval for problem 1

Answer 15

Oh ok.
But we still run into that issue, don't we?
\(\Large\rm f(3)=0\)

Answer 16

Intermediate Value Theorem
Suppose f is continuous on the interval [a,b] and L is a number between f(a) and f(b). Then there is at least one number c in (a,b) satisfying f(c)=L.

Answer 17

In this case, L is NOT between f(a) and f(b), correct?

Answer 18

I did a similar problem years ago from a textbook
Verify that IVT applies in the indicated interval and find the value of c guaranteed by the theorem.
f(x)=x^2+x-1 [0,5] f(c)=11
when I solved for x I got x=3 and x=-4
f(3) was the solution since 3 was in the interval [0,5]

Answer 19

ok so for problem 1, it does fail since f(13)=23 is not in the interval [-1,3]

Answer 20

I guess I would say that... since L is NOT in between f(a) and f(b),
the Intermediate Value Theorem can not be applied to the problem.

I don't think you even want to look for a c value.
The theorem can't be used since the condition isn't met.

Answer 21

I guess I could not visualize it and had to doublecheck that it was not in the interval. I have a hard time applying this theorem. Ok seems like I got it now. I believe I did the problem 2 and 3 correct. I am going to assume I did 4 correctly...

Answer 22

thanks for your help..... :)