Lesson 1.13 · Unit 1 · CED topic 1.16
Working with the Intermediate Value Theorem (IVT)
If you start a hike at 3,000 feet and end at 7,000 feet, you must have passed through exactly 5,000 feet at some moment, because you did not teleport. That everyday certainty is the Intermediate Value Theorem. It turns continuity, a property of a graph having no breaks, into a way to prove that an equation has a solution even when you cannot find the solution by algebra.
Learning objective.
- FUN-1.A. Explain the behavior of a function on an interval using the Intermediate Value Theorem.
Essential knowledge.
- FUN-1.A.1. If \(f\) is a continuous function on the closed interval \([a,b]\) and \(d\) is a number between \(f(a)\) and \(f(b)\), then the Intermediate Value Theorem guarantees that there is at least one number \(c\) between \(a\) and \(b\), such that \(f(c)=d\).
The theorem
Intermediate Value Theorem (IVT). If \(f\) is a continuous function on the closed interval \([a,b]\) and \(d\) is a number between \(f(a)\) and \(f(b)\), then there is at least one number \(c\) between \(a\) and \(b\) such that \(f(c)=d\).
Read the statement as a checklist. A correct use of the theorem names three things:
- \(f\) is continuous on \([a,b]\) (say why: the type of function, or a boundary check).
- \(d\) is between \(f(a)\) and \(f(b)\) (give the numbers and compare).
- Conclusion: there exists \(c\) in \((a,b)\) with \(f(c)=d\) (an existence statement).
The theorem guarantees that a solution exists. It does not say how many there are, where they are, or what the value of \(c\) is.
Showing that an equation has a solution
The most common use is a root-finding argument: to show \(f(x)=0\) has a solution in an interval, take \(d=0\) and show that \(f(a)\) and \(f(b)\) have opposite signs.
Show that \(x^3-x-1=0\) has a solution in \((1,2)\).
Let \(f(x)=x^3-x-1\). It is a polynomial, hence continuous on \([1,2]\). Compute \(f(1)=1-1-1=-1\lt0\) and \(f(2)=8-2-1=5\gt0\). Since \(0\) is between \(f(1)=-1\) and \(f(2)=5\), the Intermediate Value Theorem guarantees a number \(c\) in \((1,2)\) with \(f(c)=0\). So the equation has a solution in \((1,2)\).
Show that \(\cos x=x\) has a solution in \((0,1)\).
Rewrite as \(g(x)=\cos x-x=0\). Both \(\cos x\) and \(x\) are continuous, so \(g\) is continuous on \([0,1]\). \(g(0)=1-0=1\gt0\) and \(g(1)=\cos1-1\approx-0.4597\lt0\) (with \(x\) in radians). Zero lies between \(g(1)\) and \(g(0)\), so by the IVT there is a \(c\) in \((0,1)\) with \(g(c)=0\), that is, \(\cos c=c\).
Using a table
IVT arguments also work from a table of values, when you are told that the function is continuous.
A continuous function \(f\) on \([0,7]\) has \(f(0)=-4\), \(f(3)=2\), and \(f(7)=-1\). What does the IVT guarantee about the equation \(f(x)=0\)?
\(f(0)=-4\lt0\lt2=f(3)\), so there is at least one \(c_1\) in \((0,3)\) with \(f(c_1)=0\). And \(f(3)=2\gt0\gt-1=f(7)\), so there is at least one \(c_2\) in \((3,7)\) with \(f(c_2)=0\). These intervals do not overlap, so there are at least two solutions in \((0,7)\). The theorem does not rule out more.
When the hypotheses fail
The continuity requirement is essential. Consider \(f(x)=\dfrac1x\) on \([-1,1]\): \(f(-1)=-1\) and \(f(1)=1\), so \(0\) is between them, but \(f(x)=0\) has no solution. The IVT does not apply because \(f\) is not continuous on \([-1,1]\) (it is undefined at \(0\)). Similarly, if \(d\) is not between \(f(a)\) and \(f(b)\), the theorem says nothing: a continuous function with \(f(0)=-3\) and \(f(4)=5\) need not ever equal \(6\).
A related remark: because the theorem is only an existence result, you cannot use it to say a value is unique. Later, when you know a function is also increasing, you can combine the two ideas to show exactly one solution.
- IVT: \(f\) continuous on \([a,b]\) and \(d\) between \(f(a)\) and \(f(b)\) implies \(f(c)=d\) for some \(c\) in \((a,b)\).
- A complete justification states: continuity on the closed interval, the values \(f(a)\) and \(f(b)\) with \(d\) between them, and the conclusion.
- To show \(f(x)=0\) has a solution, find a sign change; to solve \(g(x)=h(x)\), apply the theorem to \(g-h\).
- It proves existence only; it does not locate \(c\) or count the solutions.
- If \(f\) is not continuous on the whole interval, or \(d\) is outside the range between the endpoint values, the theorem does not apply.
- Show that \(x^3+x-3=0\) has a solution in \((1,2)\).
Show answer
\(f(x)=x^3+x-3\) is continuous (a polynomial). \(f(1)=-1\lt0\) and \(f(2)=7\gt0\). By the IVT there is a \(c\) in \((1,2)\) with \(f(c)=0\).
- A function \(f\) is continuous on \([2,6]\) with \(f(2)=5\) and \(f(6)=1\). Must \(f(x)=3\) have a solution in \((2,6)\)? Justify.
Show answer
Yes. \(f\) is continuous on \([2,6]\) and \(3\) is between \(f(6)=1\) and \(f(2)=5\), so the IVT guarantees a \(c\) in \((2,6)\) with \(f(c)=3\).
- Explain why the IVT does not show that \(f(x)=\dfrac1x\) has a zero in \([-1,1]\), even though \(f(-1)\lt0\lt f(1)\).
Show answer
\(f\) is not continuous on \([-1,1]\) because it is not defined at \(x=0\). The hypothesis fails, and indeed \(\dfrac1x=0\) has no solution.
- Show that \(e^x+x-3=0\) has a solution in \((0,1)\).
Show answer
\(h(x)=e^x+x-3\) is continuous. \(h(0)=1+0-3=-2\lt0\) and \(h(1)=e-2\approx0.718\gt0\). By the IVT there is a \(c\) in \((0,1)\) with \(h(c)=0\).
- A continuous function has \(f(0)=-3\) and \(f(4)=5\). Does the IVT guarantee a solution of \(f(x)=6\) in \((0,4)\)?
Show answer
No. The number 6 is not between \(f(0)=-3\) and \(f(4)=5\), so the theorem does not apply. (The function might never reach 6.)
- At 8 a.m. the temperature is \(54^\circ\)F; at 2 p.m. it is \(71^\circ\)F. Assuming the temperature varies continuously, explain why it was exactly \(60^\circ\)F at some time between.
Show answer
The temperature function is continuous on \([8,14]\) (hours), and \(60\) is between \(54\) and \(71\). By the IVT there is a time \(t\) in \((8,14)\) when the temperature was \(60^\circ\)F.