Posted by **sonya** on Thursday, January 24, 2008 at 6:32pm.

x^3-4x-2 [2,3]

we have to use the intermediate value thereom. not sure how to do this.

- math -
**Damon**, Thursday, January 24, 2008 at 7:30pm
x^3-4x-2 [2,3]

the function is continuous between 2 and 3

at 2 f(2) = -2

at 3 f(3) = +13

The theorem states that for every number N between -2 and +13, there is some x between 2 and 3 such that f(x) = N

Now they did not give you an N so I will say 5 for example is between -2 and +13

The theorem says that some x between 2 and 3 will come out with f(x)= 5

I would have to solve it by trial and error but I know to start between 2 and 5

x f(x)

2 -2

try 2.5 --> 3.63 too small

try 2.7 --> 6.88 too big

try 2.6 --> 5.18 closer, a bit big

try 2.58 -> 4.85 a little small

try 2.585-> 4.93 well you get the idea

3 13

The theorem says that

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