# Thread: I don't understand the Intermediate Value Theorem?

1. ## I don't understand the Intermediate Value Theorem?

Let f be continuous on the closed interval [-2, 5]. If f(-2) = 3 and f(5) = -7, then the IVt guarentees that:

a) 7≤f(x)≤3 for all x between -2 and 5
b) f'(c) = -10/7 for at least one c between [-2, 5]
c) f(c) = -3 for at least one c between [-2, 5]
d) f(c) = 0 for at least one c between [-7, 3]
e) f(x) is continually decreasing between [-2, 5]

I don't get how to apply the IVT theorem.. My AP exam is in a little over a week. HELP!!!

2. all the IVT is saying is that if f(x) is cont on [a,b]

it takes on all values between f(a) and f(b)

3. So then would the answer be E?

4. Originally Posted by elpermic
So then would the answer be E?
All you know is the f is continuous on the interval. Why couldn't f decrease then increase and then decrease again as we moved from $\displaystyle x=-2 \to 5$?

5. Hint f takes on all values between -7 and 3

6. Originally Posted by danny arrigo
All you know is the f is continuous on the interval. Why couldn't f decrease then increase and then decrease again as we moved from $\displaystyle x=-2 \to 5$?

So then the answer would be A?

7. Originally Posted by elpermic
So then the answer would be A?
I'm guessing that (a) has a typo and should read

$\displaystyle -7 \le f(x) \le 3$

but nothing says that $\displaystyle f(x)$ can be greater than 3 or less than -7. Look at Cal26's second post.

8. try c and d also then we will have hit every wrong answer

sorry for the sarcasm but -10/7 is the only number between -7 and 3

9. So the answer is B

How?

10. Apologies didn't see the frime on f in b

and I was reading 3 on c

Co c is the answer since -3 is between -7 and 3

that;s what i get for sarcasm