# math

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

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

• math -

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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