Expert: Paul Klarreich - 12/15/2006

Question
How do I use the Intermediate Value Theorum to show that a function has at least one solution. For example, how do I prove that x^3 + 2x = 10 has at least one solution?

Answer
Questioner:  Felicia
Category:  Calculus
 
Question:  How do I use the Intermediate Value Theorem to show that a function has at least one solution. For example, how do I prove that x^3 + 2x = 10 has at least one solution?

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Hi, Felicia,

There is no such thing as a solution to a function, but there could be a solution to an equation.  And, of course, not every equation has a solution.  

You will not be able to prove using the IVT that  x^2 + 4 = 0 has a (real) solution.

So you must mean "show that a given EQUATION has at least one solution."  
That's related to "show that a given FUNCTION has at least one real ZERO."  

A zero of a function is simply a value of x such that f(x) = 0, which means a value of x where the graph crosses the x-axis.

Now to your example

x^3 + 2x = 10

This equation is equivalent to:

x^3 + 2x - 10 = 0

So let f(x) be the left side, f(x) = x^3 + 2x - 10,  and try to prove that f(x) has a zero.  That is, prove that for some x,  f(x) = 0.

Now your IVT says that if  f(x) is continuous on [a,b] then f(x) must take on every value between f(a) and f(b).

HERE'S THE CLUE.  Find values of a and b, presumably with  a < b, such that:

f(a) is positive.
f(b) is negative.

OR vice versa.  It's enough that f(a) and f(b) have opposite signs.  Oh, and don't forget to say why f(x) is continuous.  In this case it's a polynomial and those are continuous everywhere -- that's why we love them.

Well, if f(a) and f(b) have opposite signs, say f(a) is +, f(b) is -, then f(x) takes on all values between your + and your -.

But zero is between any + and any -, so f(x) must be zero somewhere in [a,b].

Now YOUR f(x) = x^3 + 2x - 10.

How about a value where f(x) is negative?  

Looks like  f(0) = - 10,  so a can be 0.

Now pick a nice large positive number.

f(10) = 1000 + 20 - 10 = 1010.  That's +, so b can be 10.

Therefore, there exists a number c, between 0 and 10 such that f(c) = 0.

Since  c^3 + 2c + 10 = 0,
      c^3 + 2c = -10, and
 c is a solution to x^3 + 2x = 10