Justify with the Intermediate Value Theorem

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continuous on \([a,b]\); k between \(f(a)\) and \(f(b)\)

then f(c) = k for some c between a and b

What to Remember

Write the hypotheses before the conclusion

The Intermediate Value Theorem needs a function that is continuous on the closed interval \([a, b]\), and a number k between \(f(a)\) and \(f(b)\).

Then there is at least one c in \((a, b)\) with \(f(c) = k\), when k is strictly between the endpoint values.

Between does not mean anywhere

If k is not between \(f(a)\) and \(f(b)\), the theorem does not guarantee \(f(c) = k\).

Continuity has to cover the whole interval

A hole, a vertical asymptote, or a missing endpoint means the function is not continuous on that closed interval. The theorem does not apply, even if a simpler formula would hit the value.

At least one, not exactly one

The theorem guarantees existence. It does not say the solution is unique, and it does not name the solution.

Level 1

Easy

The function is already known to be continuous. Decide whether k sits between the endpoint values.

Example

f is continuous on \([2, 6]\), \(f(2) = -3\), and \(f(6) = 5\). Explain why there is a c in \((2, 6)\) with \(f(c) = 1\).

f is continuous on \([2, 6]\).

\(f(2) = -3 \lt 1 \lt 5 = f(6)\)

1 is between the endpoint values.

By the Intermediate Value Theorem, there is a c in \((2, 6)\) with \(f(c) = 1\).

Answer: yes, \(f(c) = 1\) for some c in \((2, 6)\)

f is continuous on \([0, 4]\), \(f(0) = 1\), and \(f(4) = 7\). Does the theorem guarantee \(f(c) = 4\) for some c in \((0, 4)\)?

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f is continuous on \([0, 4]\).

\(1 \lt 4 \lt 7\)

By the Intermediate Value Theorem, yes.

Answer: yes

Same f. Does the theorem guarantee \(f(c) = 0\) for some c in \((0, 4)\)?

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f is continuous on \([0, 4]\).

The endpoint values are 1 and 7.

0 is not between 1 and 7.

The theorem does not guarantee a zero.

Answer: no

f is continuous on \([-2, 3]\), \(f(-2) = 8\), and \(f(3) = -1\). Does the theorem guarantee a zero in \((-2, 3)\)?

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f is continuous on \([-2, 3]\).

\(-1 \lt 0 \lt 8\)

0 is between the endpoint values.

By the Intermediate Value Theorem, yes.

Answer: yes

f is continuous on \([0, 4]\) and \(f(0) = f(4) = 2\). Must there be a c in \((0, 4)\) with \(f(c) = 2\)?

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The theorem guarantees \(f(c) = 2\) somewhere on the closed interval.

The endpoints already satisfy that.

A function may leave 2 on the open interval and return only at the other endpoint.

An interior point is not guaranteed.

Answer: no

Let \(g(x) = x^3 + x - 5\). Explain why \(x^3 + x = 5\) has a solution in \((1, 2)\).

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g is a polynomial, so g is continuous on \([1, 2]\).

\(g(1) = 1 + 1 - 5 = -3\)

\(g(2) = 8 + 2 - 5 = 5\)

\(-3 \lt 0 \lt 5\)

By the Intermediate Value Theorem, \(g(c) = 0\) for some c in \((1, 2)\).

That c is a solution of \(x^3 + x = 5\).

Answer: yes, a solution in \((1, 2)\)

Level 2

Medium

Check continuity before you use the theorem. Then pick the statement that must be true.

Example

f is continuous on \([0, 4]\), \(f(0) = 2\), and \(f(4) = 9\). Which must be true?

A. \(f(c) = 0\) for some c in \((0, 4)\)

B. \(f(c) = 6\) for some c in \((0, 4)\)

C. f is differentiable on \((0, 4)\)

D. \(f(2) = 6\)

f is continuous on \([0, 4]\).

\(2 \lt 6 \lt 9\)

By the Intermediate Value Theorem, \(f(c) = 6\) for some c in \((0, 4)\).

0 is not between 2 and 9.

Continuity does not give differentiability, and it does not give the value at \(x = 2\).

Answer: B. \(f(c) = 6\) for some c in \((0, 4)\)

Does the Intermediate Value Theorem guarantee that \(f(x) = \dfrac{1}{x}\) hits 5 somewhere on \([0, 1]\)?

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The theorem needs continuity on the closed interval.

\(f\) is not defined at \(x = 0\).

f is not continuous on \([0, 1]\).

The theorem does not apply.

Answer: no

Does the Intermediate Value Theorem apply to \(h(x) = \dfrac{x^2 - 4}{x - 2}\) on \([1, 3]\)?

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\(h\) is not defined at \(x = 2\).

2 is inside \([1, 3]\).

h is not continuous on \([1, 3]\).

The theorem does not apply. Canceling \(x - 2\) produces a different function.

Answer: no

f is continuous on \([-1, 1]\), \(f(-1) = -4\), and \(f(1) = 3\). How many zeros does the theorem guarantee?

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\(-4 \lt 0 \lt 3\)

The theorem guarantees at least one zero in \((-1, 1)\).

It does not say the zero is unique.

Answer: at least one

Let \(p(x) = e^x - 4\). Explain why \(e^x = 4\) has a solution in \((1, 2)\).

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p is continuous on \([1, 2]\).

\(e \lt 3 \lt 4\), so \(p(1) = e - 4 \lt 0\).

\(e > 2\), so \(e^2 > 4\) and \(p(2) = e^2 - 4 > 0\).

\(p(1) \lt 0 \lt p(2)\)

By the Intermediate Value Theorem, \(p(c) = 0\) for some c in \((1, 2)\).

Answer: yes, a solution in \((1, 2)\)

f is continuous on \([3, 8]\) and \(f(3) = f(8) = 5\). Which must be true?

A. f has a zero in \((3, 8)\)

B. \(f(c) = 5\) for some c in \((3, 8)\)

C. \(f(c) = 6\) for some c in \((3, 8)\)

D. none of these must be true

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0 and 6 are not between 5 and 5.

The value 5 is already taken at the endpoints.

f can leave 5 on the open interval and come back only at \(x = 8\).

None of A, B, or C must be true.

Answer: D. none of these must be true

Level 3

Hard

Write the full justification, or explain why a reason is not valid.

Example

Let \(f(x) = x^3 - 4x - 1\). Show that the equation \(x^3 - 4x - 1 = 0\) has a solution in \((2, 3)\).

f is a polynomial, so f is continuous on \([2, 3]\).

\(f(2) = 8 - 8 - 1 = -1\)

\(f(3) = 27 - 12 - 1 = 14\)

\(-1 \lt 0 \lt 14\)

By the Intermediate Value Theorem, there is a c in \((2, 3)\) with \(f(c) = 0\).

Answer: a solution in \((2, 3)\)

f is continuous on \([1, 5]\), \(f(1) = 4\), and \(f(5) = 9\). A student says f has a zero because the endpoint values are positive. Is that a valid use of the theorem?

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The endpoint values are 4 and 9.

0 is not between 4 and 9.

The Intermediate Value Theorem does not guarantee a zero.

Positive endpoint values are not a reason for a root.

Answer: no

Let \(f(x) = \dfrac{x - 1}{x - 1}\) for \(x \neq 1\), and suppose f is not defined at 1. Does the Intermediate Value Theorem guarantee \(f(c) = 1\) for some c in \((0, 2)\)?

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f is not defined at \(x = 1\), which is inside \([0, 2]\).

f is not continuous on \([0, 2]\).

The theorem does not apply, so it does not guarantee the value.

For \(x \neq 1\), \(f(x) = 1\), so the value happens to be hit. That fact is not the Intermediate Value Theorem.

Answer: no; the theorem does not apply

f is continuous on \([0, 2]\) and on \([3, 5]\). Also \(f(0) = -1\) and \(f(5) = 2\). Does the theorem on \([0, 5]\) guarantee a zero?

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The theorem needs continuity on the whole closed interval \([0, 5]\).

Continuity on two separate pieces does not give continuity on \([0, 5]\).

Nothing is known about \((2, 3)\).

The theorem does not apply on \([0, 5]\).

Answer: no

f is continuous on \([-4, -1]\), \(f(-4) = 6\), and \(f(-1) = -2\). Which values does the theorem guarantee on \((-4, -1)\): 0, 10, or both?

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f is continuous on \([-4, -1]\).

\(-2 \lt 0 \lt 6\), so 0 is guaranteed.

10 is not between \(-2\) and 6.

10 is not guaranteed.

Answer: 0 only

f is continuous on \([1, 4]\). \(f(1) = -6\) and \(f(4) = 2\). A second continuous function g satisfies \(g(1) = 2\) and \(g(4) = -6\). Explain why \(f(x) = g(x)\) has a solution in \((1, 4)\).

Show Answer

Let \(d(x) = f(x) - g(x)\).

f and g are continuous, so d is continuous on \([1, 4]\).

\(d(1) = -6 - 2 = -8\)

\(d(4) = 2 - (-6) = 8\)

\(-8 \lt 0 \lt 8\)

By the Intermediate Value Theorem, \(d(c) = 0\) for some c in \((1, 4)\).

Then \(f(c) = g(c)\).

Answer: yes, \(f(c) = g(c)\) for some c in \((1, 4)\)

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