- Compare the target output with the two endpoint outputs. - Separate what the theorem says about existence from any claim about the number of solutions.

1. The function is continuous on the closed interval \([2, 7]\). 2. The target value \(10\) lies between the endpoint values \(5\) and \(13\). 3. The Intermediate Value Theorem guarantees at least one \(c\in(2, 7)\) such that \(g(c)=10\). 4. The theorem gives existence, not uniqueness, so it does not rule out several such inputs.

There is at least one \(c\in(2, 7)\) with \(g(c)=10\). The Intermediate Value Theorem does not guarantee that this solution is unique.

- Identify the input interval and the two endpoint outputs. - Check whether the target temperature lies between those outputs.

1. The temperature function is continuous over the one-hour interval. 2. The value \(400\,\text{°F}\) lies between \(480\,\text{°F}\) and \(350\,\text{°F}\). 3. By the Intermediate Value Theorem, there is at least one time between \(1{:}00\) p.m. and \(2{:}00\) p.m. when \(T(t)=400\,\text{°F}\). 4. The theorem does not determine the exact time or guarantee that the temperature is reached only once.

The sample must be at \(400\,\text{°F}\) at least once between \(1{:}00\) p.m. and \(2{:}00\) p.m.

- Read the function values at the two ends of the interval. - Is the target height between those values? - What does continuity guarantee about the heights reached in between?

1. From the curve, \(f(-2)=-6\) and \(f(2)=6\). 2. The target value \(4\) lies strictly between the endpoint heights. 3. The graph is continuous on the closed interval, so the Intermediate Value Theorem guarantees at least one \(c\in(-2,2)\) with \(f(c)=4\).

Yes. The curve is continuous and the endpoint heights \(-6\) and \(6\) lie on opposite sides of \(4\).

- Follow the graph between the two stated endpoints. - Does the curve pass through all heights between them? - Check the continuity condition of the theorem.

1. The graph jumps from height \(-1\) on the left to \(1\) on the right at \(x=0\). 2. It never reaches height \(0\) anywhere between \(-1\) and \(1\). 3. The Intermediate Value Theorem requires continuity on the interval. That condition fails, so opposite-sign endpoint values alone do not guarantee a zero.

The function has a jump and is not continuous on \([-1,1]\). The graph never reaches \(0\), so IVT does not apply.

- Check the condition on the function over the entire closed interval before comparing endpoint values. - Compare the desired output with the outputs at the two endpoints.

1. The function \(f\) is a polynomial, so it is continuous on \([0, 1]\). 2. Evaluate the endpoints: \(f(0)=-1\) and \(f(1)=1\). 3. The target value \(0\) lies between \(-1\) and \(1\). 4. By the Intermediate Value Theorem, there is at least one \(c\in(0, 1)\) such that \(f(c)=0\).

Because \(f\) is continuous on \([0, 1]\), with \(f(0)=-1<0<1=f(1)\), the Intermediate Value Theorem guarantees at least one \(c\in(0, 1)\) with \(f(c)=0\).

- Checking only the endpoint values is not enough; verify every hypothesis of the theorem. - Look at what output values the piecewise function can actually take.

1. The endpoint values do straddle \(0\): \(h(-1)=-1\) and \(h(1)=1\). 2. However, \(h\) has a jump discontinuity at \(x=0\), so it is not continuous on \([-1,1]\). 3. Therefore the Intermediate Value Theorem does not apply. 4. In fact, the function takes only the values \(-1\) and \(1\), so there is no input for which \(h(x)=0\).

The claim is false. The function is not continuous on \([-1,1]\), so the Intermediate Value Theorem cannot be used, and \(h(x)=0\) has no solution there.

- Rewrite the equation so that finding a solution becomes finding a zero of one continuous function. - Evaluate that function at the endpoints and compare the signs.

1. Define \(F(x)=\cos x-x\). Both cosine and the identity function are continuous, so \(F\) is continuous on \([0, 1]\). 2. At the endpoints, \(F(0)=1\) and \(F(1)=\cos 1-1\approx-0.4597\). 3. Since \(0\) lies between these endpoint values, the Intermediate Value Theorem gives at least one \(c\in(0, 1)\) with \(F(c)=0\). 4. The equation \(F(c)=0\) is equivalent to \(\cos c=c\).

Let \(F(x)=\cos x-x\). It is continuous on \([0, 1]\), and \(F(0)=1>0\) while \(F(1)=\cos 1-1\approx-0.4597<0\). Therefore the Intermediate Value Theorem guarantees at least one solution to \(\cos x=x\) in \((0, 1)\).

- Write the interval of output values lying between the two endpoint outputs. - The theorem guarantees every target inside that output interval, not targets beyond it.

1. Because \(f\) is continuous on \([-2, 3]\), it must attain every value between the endpoint outputs \(-4\) and \(7\). 2. The values \(-3\), \(0\), and \(6\) all lie between \(-4\) and \(7\), so each is guaranteed to occur at least once. 3. The value \(8\) lies outside the interval of endpoint values, so the Intermediate Value Theorem gives no guarantee that \(f(c)=8\) anywhere on \([-2, 3]\).

Guaranteed: \(-3\), \(0\), and \(6\). Not guaranteed by the Intermediate Value Theorem: \(8\).

- Evaluate the function at the midpoint rather than drawing a conclusion from the endpoints alone. - Distinguish a condition that guarantees a root from a condition that is necessary for a root.

1. The function \(f(x)=(x-1)^2\) is a polynomial, so it is continuous on \([0,2]\). 2. Its endpoint values are \(f(0)=1\) and \(f(2)=1\), so there is no endpoint sign change. 3. Nevertheless, \(f(1)=0\), so the function does have a zero between the endpoints. 4. An endpoint sign change is a sufficient condition for using the Intermediate Value Theorem to guarantee a zero, but the absence of a sign change does not prove that no zero exists.

The statement is false. Although \(f(0)=f(2)=1>0\), \(f(1)=0\). A sign change lets the Intermediate Value Theorem guarantee a zero, but no sign change does not rule one out.

- Compare the midpoint value with each original endpoint value. - Keep the subinterval whose endpoint outputs straddle zero.

1. On \([2, 3]\), \(f(2)=-3\) and \(f(3)=1\) have opposite signs. 2. Because \(f\) is continuous, the Intermediate Value Theorem guarantees at least one zero in \((2, 3)\). 3. On \([3, 4]\), both given endpoint values are positive, so the theorem does not guarantee a zero in that half from these data.

The guaranteed half-interval is \((2, 3)\).

- Notice whether the target lies strictly between the endpoint outputs or equals one of them. - Separate “attained somewhere on the closed interval” from “attained at an interior input.”

1. The value \(3\) is attained at the endpoint \(x=0\). 2. The Intermediate Value Theorem guarantees that every value between \(3\) and \(8\) is attained somewhere on the closed interval \([0, 4]\). 3. For a target strictly between the endpoint values, the theorem guarantees an interior input. But the target \(3\) equals an endpoint value. 4. Therefore the theorem does not guarantee another input in \((0, 4)\) with \(f(c)=3\).

No. The theorem guarantees \(f(0)=3\) on the closed interval, but it does not guarantee a second occurrence of \(3\) at an interior point.

- First use continuity and the endpoint signs only to state the theorem's guarantee. - Then analyze the particular function separately to see whether more than one zero occurs.

1. The polynomial \(f\) is continuous on \([-2, 2]\). 2. The endpoint values are \(f(-2)=-6\) and \(f(2)=6\), so the Intermediate Value Theorem guarantees at least one zero in \((-2, 2)\). 3. Factor \(f(x)=x(x-1)(x+1)\). 4. The actual zeros are \(x=-1\), \(x=0\), and \(x=1\). Thus the theorem's existence guarantee does not specify the total number of zeros.

The Intermediate Value Theorem guarantees at least one zero, but \(f\) actually has three zeros on the interval: \(-1\), \(0\), and \(1\).

- Look for pairs of known inputs whose outputs have opposite signs. - Among all such pairs, compare the widths of their input intervals.

1. A zero is guaranteed whenever two known endpoint values have opposite signs. 2. The values \(F(1.25)=-0.1\) and \(F(1.375)=0.05\) have opposite signs. 3. Therefore the Intermediate Value Theorem guarantees a zero in \((1.25, 1.375)\). 4. This interval is narrower than the other sign-changing intervals available from the given data.

A zero is guaranteed in \((1.25, 1.375)\).

- Evaluate the polynomial at the three integer inputs that divide the two requested intervals. - Treat each interval as a separate application of the theorem.

1. The function \(f\) is a polynomial, so it is continuous on both \([0, 1]\) and \([1, 2]\). 2. Evaluate \(f(0)=1\), \(f(1)=-2\), and \(f(2)=9\). 3. Since the signs change from \(f(0)>0\) to \(f(1)<0\), the Intermediate Value Theorem guarantees a zero in \((0, 1)\). 4. Since the signs change again from \(f(1)<0\) to \(f(2)>0\), the theorem guarantees another zero in \((1, 2)\). The intervals are disjoint, so the zeros are distinct.

At least two distinct real zeros are guaranteed: one in \((0, 1)\) and one in \((1, 2)\).

- For a rational function, first check that its denominator never vanishes on the chosen interval. - Then compare the requested output with the two endpoint outputs.

1. The denominator \(x-2\) is nonzero throughout \([3, 5]\), so the rational function \(f\) is continuous on that closed interval. 2. The endpoint values are \(f(3)=4\) and \(f(5)=2\). 3. The target value \(3\) lies between \(4\) and \(2\). 4. By the Intermediate Value Theorem, there is at least one \(c\in(3, 5)\) such that \(f(c)=3\).

A solution is guaranteed in \((3, 5)\) because \(f\) is continuous on \([3, 5]\) and \(f(3)=4>3>2=f(5)\).

- The continuity hypothesis is already supplied; read only the endpoint signs needed for each interval. - Opposite endpoint signs place \(0\) between the two endpoint outputs. - Same-sign endpoints mean “not guaranteed,” not “impossible.”

1. The plotted endpoint values for \([-2,-1]\) have opposite signs, so continuity and the Intermediate Value Theorem guarantee at least one zero in \((-2,-1)\). 2. The plotted endpoint values for \([-1,0]\) are both positive. The Intermediate Value Theorem therefore does not guarantee a zero in \((-1,0)\). 3. The plotted endpoint values for \([1,2]\) have opposite signs, so the Intermediate Value Theorem guarantees at least one zero in \((1,2)\).

a) Yes; IVT guarantees a zero in \((-2,-1)\). b) No; these endpoint values do not guarantee a zero. c) Yes; IVT guarantees a zero in \((1,2)\).

- Read the heights at the stated endpoints. - Is the target output between the endpoint values? - IVT requires continuity, not a curve that rises steadily.

1. The graph has \(f(0)=2\) and \(f(3)=5\). 2. Since \(4\) lies between \(2\) and \(5\), and the function is continuous on the interval, IVT guarantees a point with \(f(c)=4\). 3. The curve's dip does not change this guarantee; it need not rise throughout the interval.

Yes. The continuous curve starts at height \(2\) and ends at \(5\), so it must reach \(4\).

- Check the whole interval for intersections with the horizontal level \(y=1\). - Compare the endpoint heights on each half of the interval. - Distinguish “at least one” from “exactly one.”

1. The curve intersects the height \(1\) twice, at \(x=-1\) and \(x=1\). 2. On \([-2,0]\), endpoint heights are \(4\) and \(0\), so IVT guarantees at least one point at height \(1\). 3. On \([0,2]\), endpoint heights are \(0\) and \(4\), so IVT guarantees at least one more such point. 4. IVT guarantees existence within each interval but does not by itself rule out additional intersections.

Two points, at \(x=-1\) and \(x=1\). IVT guarantees at least one in each half-interval, not uniqueness.

- Read the curve's height at both ends of each listed interval. - Look for one endpoint below the target and the other above it. - Separate what the endpoints guarantee from what the rest of the graph reveals.

1. At the endpoints of A, \(f(-2)=-2\) and \(f(0)=0\), both below \(1\). Those values alone do not guarantee reaching \(1\). 2. At the endpoints of B, \(f(0)=0\) and \(f(2)=2\), which lie on opposite sides of \(1\). 3. At the endpoints of C, \(f(2)=2\) and \(f(2.5)=8.125\), both above \(1\). 4. The curve is continuous, so IVT guarantees at least one solution inside interval B. The graph also rises above \(1\) and falls back below it inside interval A, but that extra information is not contained in A's endpoint values.

B, \([0,2]\), because its endpoint values \(0\) and \(2\) surround \(1\). Interval A does reach \(1\), but its two endpoint values are both below \(1\), so they alone do not guarantee a crossing.

- Look for separate subintervals whose endpoint values have opposite signs. - To count distinct guaranteed zeros, check whether the intervals containing them overlap.

1. On \([-3, -1]\), the endpoint values have opposite signs, so there is at least one zero in \((-3, -1)\). 2. On \([-1, 1]\), the endpoint values again have opposite signs, so there is at least one zero in \((-1, 1)\). 3. On \([1, 3]\), the endpoint values have opposite signs, so there is at least one zero in \((1, 3)\). 4. These three open intervals are disjoint, so the guaranteed zeros are distinct. Therefore at least three distinct zeros are guaranteed.

At least \(3\) distinct zeros are guaranteed: one in each of \((-3, -1)\), \((-1, 1)\), and \((1, 3)\).

- Recall exactly which interval must satisfy the continuity hypothesis. - Ask whether separately assigned endpoint values must match nearby values when endpoint continuity is not given.

1. The Intermediate Value Theorem requires continuity on the entire closed interval \([0, 1]\), including continuity at the endpoints from within the interval. 2. Continuity only on \((0, 1)\) does not connect the separately assigned endpoint values to nearby function values. 3. For example, a function could equal \(1\) for every \(0<x<1\), while still being assigned \(f(0)=-2\) and \(f(1)=2\). 4. That example has no zero in \((0, 1)\), so the stated information is insufficient to apply the theorem.

No. Continuity on \((0, 1)\) alone is insufficient; the theorem requires continuity on \([0, 1]\).

- First make the two pieces agree at their joining input. - After continuity is established, compare the requested output with the endpoint outputs of the full interval.

1. The only possible break is at \(x=1\). The left branch gives \(f(1)=3\), while the right-hand limit is \(k\). 2. Continuity requires \(k=3\). 3. With \(k=3\), the function is continuous on \([0, 2]\), with \(f(0)=2\) and \(f(2)=6\). 4. Since \(4\) lies between \(2\) and \(6\), the Intermediate Value Theorem guarantees some \(c\in(0, 2)\) with \(f(c)=4\).

\(k=3\). Then \(f\) is continuous on \([0, 2]\), and because \(2=f(0)<4<f(2)=6\), the Intermediate Value Theorem guarantees at least one \(c\in(0, 2)\) with \(f(c)=4\).

- Suppose the endpoint signs were opposite and ask what value would lie between them. - Use the theorem to compare that consequence with the given fact that no zero exists.

1. Assume, for contradiction, that \(f(a)\) and \(f(b)\) have opposite signs. 2. Then \(0\) lies strictly between the endpoint values. 3. Since \(f\) is continuous on \([a, b]\), the Intermediate Value Theorem would guarantee some \(c\in(a, b)\) with \(f(c)=0\). 4. This contradicts the statement that \(f\) has no zeros on \([a, b]\). Therefore the endpoint values cannot have opposite signs.

They cannot have opposite signs. If they did, continuity and the Intermediate Value Theorem would force a zero between \(a\) and \(b\), contradicting the hypothesis.

- Endpoint values are useful only after checking continuity over the entire closed interval. - Simplify the rational expression while keeping its original domain restriction.

1. The target value \(2\) lies between the endpoint values \(1\) and \(3\). 2. However, the original function is undefined at \(x=1\), so it is not continuous on \([0, 2]\). The Intermediate Value Theorem does not apply. 3. For \(x\ne1\), the expression simplifies to \(f(x)=x+1\). 4. The equation \(x+1=2\) has the only solution \(x=1\), but that input is excluded from the domain. Therefore the original function never attains the value \(2\) on \((0, 2)\).

No. The function is not continuous on \([0, 2]\) because it is undefined at \(x=1\). In fact, \(f(x)=2\) would require \(x=1\), which is not in the domain, so there is no solution.

- Use the endpoint values first to establish existence. - Then ask what strict increase implies about two inputs having the same output.

1. Because \(f\) is continuous on \([-1, 4]\) and \(-2<0<5\), the Intermediate Value Theorem guarantees at least one \(c\in(-1, 4)\) with \(f(c)=0\). 2. Since \(f\) is strictly increasing, two different inputs cannot have the same output. 3. Therefore there can be at most one zero. 4. Combining existence from the Intermediate Value Theorem with uniqueness from strict increase gives exactly one solution.

Exactly one solution exists. The Intermediate Value Theorem gives at least one zero; strict increase makes that zero unique.

- Express both endpoint outputs in terms of the parameter. - Ask when zero lies between those two outputs, allowing an endpoint itself to be zero.

1. Each \(f_k\) is a polynomial, so it is continuous on \([-1, 0]\). 2. The endpoint values are \(f_k(-1)=k+3\) and \(f_k(0)=k\). 3. A zero is guaranteed from the endpoint data when \(0\) lies between these values, including equality at an endpoint. Thus \(k(k+3)\le0\). 4. This inequality holds for \(-3\le k\le0\).

\(k\in[-3, 0]\).