# I don't understand the Intermediate Value Theorem?

• Apr 26th 2009, 04:23 PM
elpermic
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!!!
• Apr 26th 2009, 04:25 PM
Calculus26
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)
• Apr 26th 2009, 04:32 PM
elpermic
So then would the answer be E?
• Apr 26th 2009, 04:37 PM
Jester
Quote:

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 $x=-2 \to 5$?
• Apr 26th 2009, 04:43 PM
Calculus26
Hint f takes on all values between -7 and 3
• Apr 26th 2009, 04:43 PM
elpermic
Quote:

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 $x=-2 \to 5$?

So then the answer would be A?
• Apr 26th 2009, 04:47 PM
Jester
Quote:

Originally Posted by elpermic
So then the answer would be A?

I'm guessing that (a) has a typo and should read

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

but nothing says that $f(x)$ can be greater than 3 or less than -7. Look at Cal26's second post.
• Apr 26th 2009, 04:48 PM
Calculus26
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
• Apr 26th 2009, 04:51 PM
elpermic