Intermediate Value Theorem

Reading

• Sections 2.8

Practice problems

• Section 2.8: 1, 5, 7, 19
• In class: 2.8 14: \(\tan x = x\) has infinitely many solutions
• To turn in (due Wednesday, together with 2.7): 2.8 2, 6

Notes

Intermediate Value Theorem

Intermediate Value Theorem

If \(f(x)\) is a continuous function on \([a,b]\) and \(L\) is a number between \(f(a)\) and \(f(b)\), then there must be a \(c\in[a,b]\) such that \(f(c) = L\).

Simply put: A continuous function takes all values between \(f(a)\) and \(f(b)\).

Important special case: If \(f(a)\) and \(f(b)\) have opposite signs, then the function \(f\) must have a zero somewhere between \(a\) and \(b\).

• Make sure to have a good graphical representation of this result.
• The importance of the theorem is that it allows us to guaranteed the existence of solutions to equations.
• Example: We will show that \(\sqrt{2}\) exists:
  • The square root is a number \(c\) such that \(c^2 = 2\).
  • Consider the function \(f(x) = x^2\) on the interval \([1,2]\).
  • Then \(f(1) < 2 < f(2)\).
  • The IVT tells us there must be a \(c\in[1,2]\) with \(f(c) = 2\).
• Example 2: Show that there is a positive \(x\) with \(x = \sin x\).
  • Idea: Bring everything to one side: \(f(x) = x - \sin x\).
  • Look for a zero.
• Bisection method: Allows us to zero in on a solution:
  • In the previous example, consider \(f(1.5)\). Since it is bigger than \(2\), the \(c\) is actually in \([1,1.5]\).
  • Try \(f(1.25)\) next.
  • Keep bisecting the interval and trying the middle value.
  • Practice: Find the first 3 decimals points of \(\sqrt{2}\).