Intermediate Value Theorem

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

then \(f(c)=k\) for some \(c\)

What to Remember

The theorem

If f is continuous on \([a,b]\) and k is between \(f(a)\) and \(f(b)\), then there is at least one c in \((a,b)\) with \(f(c)=k\).

Check continuity first

IVT does not apply if there is a jump or a hole on the closed interval. Say that the function is continuous on \([a,b]\) before you name k.

Between, not equal to an endpoint only

If k is not between the endpoint values, IVT does not guarantee a solution — even if one happens to exist elsewhere.

Existence, not the x-value

IVT says a time or an x-value exists. It does not find it.

Level 1

Easy

A polynomial on a closed interval.

Example

Let \(f(x)=x^2-2x\). Is there a number c in \((1,4)\) with \(f(c)=3\)? Justify.

\(f\) is a polynomial, so it is continuous on \([1, 4]\).

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

\(f(4) = 16 - 8 = 8\)

\(-1 < 3 < 8\)

By the Intermediate Value Theorem, yes.

Answer: yes

Let \(g(x) = 2x + 1\). Is there \(c\) in \((0, 5)\) with \(g(c) = 7\)?

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Continuous on \([0, 5]\).

\(g(0) = 1\)

\(g(5) = 11\)

\(1 < 7 < 11\)

Yes.

Answer: yes

Let \(h(x) = x^3\). Is there \(c\) in \((-1, 1)\) with \(h(c) = 2\)?

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\(h(-1) = -1\)

\(h(1) = 1\)

2 is not between \(-1\) and 1.

IVT does not guarantee it. On this interval, no, because \(x^3 \leq 1\).

Answer: no

Let \(p(x) = 5 - x\). Is there \(c\) in \((1, 4)\) with \(p(c) = 2\)?

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\(p(1) = 4\)

\(p(4) = 1\)

\(1 < 2 < 4\)

Yes.

Answer: yes

Why does IVT not apply to \(f(x) = \dfrac{1}{x}\) on \([-2, 2]\) for \(k = 0\)?

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\(f\) is not continuous on \([-2, 2]\) because of the vertical asymptote at 0.

Answer: not continuous on the interval

Let \(f(x) = x^2 + 1\). Is there \(c\) in \((0, 3)\) with \(f(c) = 0\)?

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\(f(0) = 1\)

\(f(3) = 10\)

0 is not between 1 and 10.

No.

Answer: no

Level 2

Medium

Check continuity of a piecewise function, then apply IVT.

Example

Let \(Y(t) = \begin{cases} 4 + t & 0 \leq t \leq 6 \\ 16 - t & 6 < t \leq 10 \end{cases}\). Is there a time \(t\) in \((0, 10)\) with \(Y(t) = 8\)? Justify.

Left at 6: \(10\). Right at 6: \(10\). The pieces meet, so \(Y\) is continuous on \([0, 10]\).

\(Y(0) = 4\)

\(Y(10) = 6\)

8 is not between 4 and 6, so IVT on the whole interval does not apply to 8.

On \([0, 6]\): \(Y(0) = 4\) and \(Y(6) = 10\), and \(4 < 8 < 10\).

Yes, on \((0, 6)\).

Answer: yes

Let \(Y(t) = \begin{cases} 4 + t & 0 \leq t \leq 6 \\ 16 - t & 6 < t \leq 10 \end{cases}\). Is there \(t\) in \((6, 10)\) with \(Y(t) = 8\)?

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On \([6, 10]\) the right piece is continuous.

\(Y\) approaches 10 from the right of 6, and \(Y(10) = 6\).

\(6 < 8 < 10\)

Yes.

Answer: yes

Let \(f(x) = \begin{cases} x - 1 & x < 2 \\ x + 3 & x \geq 2 \end{cases}\). Does IVT guarantee \(c\) in \((0, 4)\) with \(f(c) = 3\)?

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Left at 2: 1. Right at 2: 5. Jump discontinuity.

\(f\) is not continuous on \([0, 4]\).

IVT does not apply.

Answer: no, not continuous

Let \(f(x) = x^2 - 6x + 5\). Does IVT guarantee a root in \((1, 5)\)?

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\(f\) is continuous on \([1, 5]\).

\(f(1) = 0\)

\(f(5) = 0\)

0 is an endpoint value, not a value strictly between \(f(1)\) and \(f(5)\).

\(f(x) = (x - 1)(x - 5)\), so the only roots are the endpoints.

There is no root in the open interval \((1, 5)\).

Answer: no

Let \(f(x) = x^3 - x - 1\). Show there is a root in \((1, 2)\).

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Continuous on \([1, 2]\).

\(f(1) = 1 - 1 - 1 = -1\)

\(f(2) = 8 - 2 - 1 = 5\)

\(-1 < 0 < 5\)

Yes.

Answer: yes

A temperature \(T\) is continuous on \([0, 8]\) hours with \(T(0) = 60\) and \(T(8) = 90\). Must there be a time when \(T = 75\)?

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\(60 < 75 < 90\)

Continuous on the closed interval.

Yes.

Answer: yes

Level 3

Hard

A piecewise model: limit at infinity, continuity at a join, and IVT.

Example

A mass \(Y\), in milligrams, at time \(t\) hours is\(Y(t) = \begin{cases} 6 + t & 0 \leq t \leq 5 \\ f(t) & 5 < t \leq 9 \\ 12 + 4e^{-0.3(t - 9)} & t > 9 \end{cases}\) where \(f\) is continuous and \(f(9) = 14\). Find \(\lim_{t \to \infty} Y(t)\).

For large t, use the third piece.

\(e^{-0.3(t - 9)} \to 0\)

\(12 + 4 \cdot 0\)

\(12\)

Answer: \(12\)

A mass \(Y\) is\(Y(t) = \begin{cases} 6 + t & 0 \leq t \leq 5 \\ f(t) & 5 < t \leq 9 \\ 12 + 4e^{-0.3(t - 9)} & t > 9 \end{cases}\) where \(f\) is continuous and \(f(9) = 14\). Is \(Y\) continuous at \(t = 9\)? Justify.

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\(f(9) = 14\)

Right-hand value: \(12 + 4e^{0} = 16\).

\(14 \neq 16\)

No.

Answer: no

Let \(Y(t) = 6 + t\) for \(0 \leq t \leq 5\). Is there a time \(t\) in \((0, 5)\) with \(Y(t) = 8\)? Justify.

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The first piece is continuous on \([0, 5]\).

\(Y(0) = 6\)

\(Y(5) = 11\)

\(6 < 8 < 11\)

By the Intermediate Value Theorem, yes.

Answer: yes

A mass \(Y\) is\(Y(t) = \begin{cases} 6 + t & 0 \leq t \leq 5 \\ f(t) & 5 < t \leq 9 \\ 12 + 4e^{-0.3(t - 9)} & t > 9 \end{cases}\) where \(f\) is continuous and \(f(9) = 14\). Does IVT on \([0, 9]\) guarantee a time with \(Y(t) = 13\)?

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\(Y\) is not known to be continuous at 5 unless \(\lim_{t \to 5^+} f(t) = 11\).

\(Y\) is not continuous at 9.

IVT does not apply on the whole of \([0, 9]\).

Answer: no

Explain the meaning of \(\lim_{t \to \infty} Y(t) = 12\) in context.

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In the long run, the mass approaches 12 milligrams.

Answer: the mass approaches 12 mg

A different model is \(Y(t) = 20 + 8e^{-0.5(t - 4)}\) for \(t > 4\). Find \(\lim_{t \to \infty} Y(t)\).

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\(e^{-0.5(t - 4)} \to 0\)

\(20\)

Answer: \(20\)

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