Use the Intermediate Value Theorem to show that x^2 = 2^x somewhere between 0 and 3.

Question

Use the Intermediate Value Theorem to show that x² = 2^x  somewhere between 0 and 3.

 

 

#28, practice test

Answer 1

Let f(x) = x²-2^x

 

f(x) is continuous on the interval [0,3]

 

f(0) = 0² - 2⁰ = 0 - 1 = -1 < 0

f(3) = 3² - 2³ = 9 - 8 = 1 > 0

 

Since f(0) and f(3) have opposite signs and since f(x) is a continuous function on the interval [0,3], by the Intermediate Value Theorem,

f(c) = 0 for some number c between x = 0 and x = 3.  

 

f(c) = c²-2^c = 0

 

So, c² = 2^c

 

 

 

Answer 2

In order to use the intermediate value theorem, the functions that we are focus on must be continuous.  Being that x² and 2^x are continuous on the given, we can use the intermediate value theorem.

 

According to this theorem, there must be a root in the given interval.  Make the right side of the equation equal to zero.

 

x² - 2^x = 0

 

 

If we evaluate the left side of the equation by plugging in x=0  and x=3, the results should have different signs.  Or one of the results should be zero.

 

0² - 2⁰ = 1 - 1 = 0

 

3² - 2³ = 9 - 8 = 1

 

 

The results go from 0 to 1.  One of these results is 0.  Therefore, the to functions are equal somewhere between 0 and 3.