Intermediate Value Theorem - Calculus 1

Let $P(x)$ be a polynomial of odd degree. Which of the following statements is a guaranteed consequence of the Intermediate Value Theorem?
Answer: $P(x)$ has at least one real root. A polynomial $P(x)$ of odd degree has opposite end behavior. For instance, if the leading coefficient is positive, then $\lim_{x \to \infty} P(x) = \infty$ and $\lim_{x \to -\infty} P(x) = -\infty$ . This means we can find a value $b$ such that $P(b) > 0$ and a value $a$ such that $P(a) < 0$ . Since all polynomials are continuous on $\mathbb{R}$ , the IVT guarantees that there is at least one value $c$ between $a$ and $b$ such that $P(c)=0$ . While an odd-degree polynomial has an odd number of real roots when counted with multiplicity, the IVT only guarantees the existence of one. Roots are not guaranteed to be rational.
Let $f(x)$ and $g(x)$ be functions that are continuous for all real numbers. Let $h(x) = f(g(x))$ . Suppose $g(0)=2$ , $g(1)=5$ , $f(2)=10$ , and $f(5)=-4$ . Which of the following is guaranteed by the Intermediate Value Theorem?
Answer: There exists $c \in (0, 1)$ such that $h(c) = 0$ . The composition of continuous functions is continuous, so $h(x)$ is continuous. We evaluate $h(x)$ at the endpoints of the interval $[0, 1]$ . $h(0) = f(g(0)) = f(2) = 10$ . $h(1) = f(g(1)) = f(5) = -4$ . Since $h(x)$ is continuous on $[0, 1]$ and $0$ is between $h(0)=10$ and $h(1)=-4$ , the IVT guarantees there is a $c \in (0, 1)$ such that $h(c) = 0$ . Choice B is incorrect because the domain of $h$ under consideration is based on the domain of $g$ , which is $[0, 1]$ . Choices C and D are incorrect applications of the IVT as the target values (6 and 11) are outside the range of the respective functions on the given intervals.
A function $f$ is continuous on $[0, 5]$ with selected values $f(0)=10$ , $f(3)=-2$ , and $f(5)=8$ . Which of the following is the strongest conclusion that can be drawn from one or more applications of the Intermediate Value Theorem?
Answer: There is at least one solution to $f(x)=0$ on $(0, 3)$ and at least one solution to $f(x)=0$ on $(3, 5)$ . We can apply the IVT to the sub-intervals. On $[0, 3]$ , $f(0)=10$ and $f(3)=-2$ . Since 0 is between -2 and 10, there must be a root in $(0, 3)$ . On $[3, 5]$ , $f(3)=-2$ and $f(5)=8$ . Since 0 is between -2 and 8, there must be a root in $(3, 5)$ . Therefore, there are at least two roots on $(0, 5)$ , one in each specified sub-interval. Choice B is the most specific and strongest correct conclusion. Choice A is true but weaker. Choices C and D make claims about a range of values; D is not guaranteed, and C is true but less specific than B about the locations of the roots of $f(x)=0$ .
Consider the function $f(x) = x^4 - 2x^3 - 5$ . The Intermediate Value Theorem guarantees that $f(x)$ has a root on which of the following intervals?
Answer: $[2, 3]$ As a polynomial, $f(x)$ is continuous everywhere. To guarantee a root on an interval $[a, b]$ , the IVT requires $f(a)$ and $f(b)$ to have opposite signs. We check each interval: A) $f(-1) = (-1)^4 - 2(-1)^3 - 5 = 1 + 2 - 5 = -2$ and $f(0) = -5$ . Both are negative. B) $f(0) = -5$ and $f(1) = 1 - 2 - 5 = -6$ . Both are negative. C) $f(1) = -6$ and $f(2) = 2^4 - 2(2)^3 - 5 = 16 - 16 - 5 = -5$ . Both are negative. D) $f(2) = -5$ and $f(3) = 3^4 - 2(3)^3 - 5 = 81 - 54 - 5 = 22$ . The signs are opposite. Therefore, the IVT guarantees a root on $[2, 3]$ .
Let $f(x) = e^x - x^2$ . On which of the following closed intervals does the Intermediate Value Theorem guarantee that $f(x)$ has at least one root?
Answer: $[-1, 0]$ The function $f(x) = e^x - x^2$ is continuous everywhere. To use the IVT to guarantee a root on an interval $[a, b]$ , we need $f(a)$ and $f(b)$ to have opposite signs. We check each interval: A) $f(0) = e^0 - 0^2 = 1$ and $f(1) = e^1 - 1^2 = e-1 > 0$ . Signs are the same. B) $f(-1) = e^{-1} - (-1)^2 = 1/e - 1 < 0$ and $f(0) = 1 > 0$ . Signs are opposite. C) $f(1) = e-1 > 0$ and $f(2) = e^2 - 4 > 0$ . Signs are the same. D) $f(-2) = e^{-2} - (-2)^2 = 1/e^2 - 4 < 0$ and $f(-1) = 1/e - 1 < 0$ . Signs are the same. The only interval where the function values at the endpoints have opposite signs is $[-1, 0]$ .
Let $f(x)$ be a continuous function on a closed interval $[a, b]$ such that $f(a) = f(b) = k$ . Which of the following statements is necessarily true?
Answer: The Intermediate Value Theorem cannot be used to guarantee a solution to $f(x)=N$ for any $N \neq k$ . The IVT states that for a continuous function $f$ on $[a,b]$ , for any $N$ between $f(a)$ and $f(b)$ , there is a $c \in (a,b)$ with $f(c)=N$ . Here, $f(a)=f(b)=k$ , so there are no numbers strictly between $f(a)$ and $f(b)$ . Thus, the theorem's condition for guaranteeing a solution for $f(x)=N$ is not met for any $N \neq k$ . This means the IVT is inconclusive for any $N \neq k$ ; it doesn't guarantee a solution, nor does it guarantee the absence of one (Choice A is false). The theorem's hypotheses are met, so it is applicable (Choice B is false), its conclusion is just not useful for $N \neq k$ . Choice D is false by counterexample (e.g., $f(x) = (x-c)^2 + k-1$ for some $c \in (a,b)$ ).
Let $f(x) = \frac{x^2-1}{x-1}$ . Can the Intermediate Value Theorem be used on the interval $[0, 2]$ to prove that $f(x)=N$ for some value $N$ ?
Answer: No, because the function is undefined at $x=1$ , which is in the interval $[0,2]$ . The Intermediate Value Theorem requires the function to be continuous on the entire closed interval. The function $f(x) = \frac{x^2-1}{x-1}$ has a discontinuity at $x=1$ , which is within the interval $[0, 2]$ . Because the function itself is not continuous on $[0, 2]$ , the IVT cannot be applied to $f(x)$ on this interval. While it's true that the discontinuity is removable and the theorem would apply to its continuous extension $g(x)=x+1$ (Choice C), the question specifically asks about applying the theorem to $f(x)$ , which is not possible due to the discontinuity.
A function $f(x)$ is continuous on $[0, 10]$ . It is known that $f(0) = 5$ , $f(10) = 5$ , and for some $a \in (0, 10)$ , $f(a) = 12$ . Which conclusion is guaranteed by the Intermediate Value Theorem?
Answer: There are at least two solutions to $f(x)=8$ in $(0, 10)$ . We can apply the IVT on two sub-intervals. First, consider the interval $[0, a]$ . Since $f$ is continuous, $f(0) = 5$ and $f(a) = 12$ , and 8 is between 5 and 12, there must be a solution $c_1 \in (0, a)$ to $f(x)=8$ . Second, consider the interval $[a, 10]$ . Since $f$ is continuous, $f(a) = 12$ and $f(10) = 5$ , and 8 is between 5 and 12, there must be another solution $c_2 \in (a, 10)$ to $f(x)=8$ . Since $c_1$ and $c_2$ are in non-overlapping intervals, they are distinct. Thus, there are at least two solutions in $(0, 10)$ .
Let $f$ be a continuous function on $[0, 2]$ with $f(0)=5$ and $f(2)=1$ . The Intermediate Value Theorem guarantees the existence of a number $c$ in $(0, 2)$ such that $f(c)=N$ . Which of the following is a possible value for $N$ ?
Answer: $N=\pi$ The Intermediate Value Theorem guarantees that for a continuous function $f$ on $[a,b]$ , the function takes on every value $N$ between $f(a)$ and $f(b)$ . Here, the interval is $[0, 2]$ , and the function values at the endpoints are $f(0)=5$ and $f(2)=1$ . Therefore, the IVT guarantees a solution for any $N$ such that $1 < N < 5$ . We check the options: A) 0 is not in $(1,5)$ . B) $\pi \approx 3.14159$ , which is in $(1,5)$ . C) 5 is an endpoint value; the theorem guarantees a $c$ in the open interval $(a,b)$ for values between the endpoint values. D) 6 is not in $(1,5)$ .
Let $f(x)$ and $g(x)$ be continuous functions on the interval $[a, b]$ . If $f(a) > g(a)$ and $f(b) < g(b)$ , which of the following statements must be true based on the Intermediate Value Theorem?
Answer: There is a number $c \in (a, b)$ such that $f(c) = g(c)$ . To solve this, define a new function $h(x) = f(x) - g(x)$ . Since $f(x)$ and $g(x)$ are continuous on $[a, b]$ , their difference, $h(x)$ , is also continuous on $[a, b]$ . Evaluate $h(x)$ at the endpoints: $h(a) = f(a) - g(a) > 0$ and $h(b) = f(b) - g(b) < 0$ . Because $h(x)$ is continuous and changes sign on the interval, the Intermediate Value Theorem guarantees there is at least one number $c \in (a, b)$ such that $h(c) = 0$ . This implies $f(c) - g(c) = 0$ , or $f(c) = g(c)$ .
Let $g(x) = \int_1^x (t^2 - 3) \,dt$ . What is the correct conclusion when applying the Intermediate Value Theorem to find a root of $g(x)$ on the interval $[0, 2]$ ?
Answer: The IVT guarantees a root in $(0, 2)$ because $g(0) > 0$ and $g(2) < 0$ . First, by the Fundamental Theorem of Calculus, since $t^2-3$ is a continuous function, $g(x)$ is continuous and differentiable for all $x$ . Thus, the IVT is applicable. Next, we evaluate $g(x)$ at the endpoints of $[0, 2]$ : $g(0) = \int_1^0 (t^2 - 3) \,dt = -\int_0^1 (t^2 - 3) \,dt = -[\frac{t^3}{3} - 3t]_0^1 = -(\frac{1}{3} - 3) = \frac{8}{3} > 0$ . $g(2) = \int_1^2 (t^2 - 3) \,dt = [\frac{t^3}{3} - 3t]_1^2 = (\frac{8}{3} - 6) - (\frac{1}{3} - 3) = -\frac{10}{3} - (-\frac{8}{3}) = -\frac{2}{3} < 0$ . Since $g(x)$ is continuous on $[0, 2]$ and $g(0)$ and $g(2)$ have opposite signs, the IVT guarantees at least one root in the interval $(0, 2)$ .
Let $f(x) = x + \ln(x)$ . For which of the following intervals does the Intermediate Value Theorem guarantee a solution to $f(x) = 2$ ?
Answer: $[1, 2]$ First, we must consider the domain of $f(x)$ . The natural logarithm $\ln(x)$ is defined for $x>0$ . The function $f(x)$ is continuous on its domain $(0, \infty)$ . Choice A is invalid because the function is not defined at $x=0$ . For the other intervals, we check if the value 2 is between the function values at the endpoints. B) $f(1) = 1 + \ln(1) = 1+0=1$ . $f(2) = 2 + \ln(2)$ . Since $\ln(2) > 0$ , $f(2) > 2$ . Since $f(1) < 2 < f(2)$ , the IVT guarantees a solution in $(1,2)$ . C) $f(2) = 2 + \ln(2) > 2$ and $f(3) = 3 + \ln(3) > 2$ . D) $f(e) = e + \ln(e) = e+1 > 2$ and $f(3) = 3 + \ln(3) > 2$ .
Let $f(x) = \tan(x)$ . Why can the Intermediate Value Theorem NOT be used on the interval $[0, \pi]$ to show that $f(x)=0$ has a solution, even though $f(0)=0$ and $f(\pi)=0$ ?
Answer: Because the function is not continuous on the closed interval $[0, \pi]$ . The most fundamental requirement for the IVT is that the function must be continuous on the specified closed interval. The function $f(x) = \tan(x)$ has a vertical asymptote, and is therefore discontinuous, at $x=\pi/2$ . Since $\pi/2$ is within the interval $[0, \pi]$ , the function is not continuous on this interval, and the IVT cannot be applied. The other reasons, while potentially true statements about the situation, do not represent the fundamental reason for the failure of the IVT's applicability.
Consider the function $f(x) = \begin{cases} x^2+1 & \text{if } x \le 1 \\ 3x-1 & \text{if } x > 1 \end{cases}$ . Which of the following is a correct application of the IVT for this function?
Answer: Since $f(0)=1$ and $f(2)=5$ , there must be a $c \in (0,2)$ such that $f(c)=4$ . First, we check for continuity at $x=1$ . $f(1) = 1^2+1=2$ . The limit from the right is $\lim_{x\to 1^+} (3x-1) = 2$ . Since the value and limits match, the function is continuous everywhere. Therefore, the IVT can be applied on any closed interval. Let's check the choices: A) False. On $[0,2]$ , the function values are $f(0)=1$ and $f(2)=5$ . The value $0.5$ is not between $1$ and $5$ . B) False. On $[1,2]$ , the values are $f(1)=2$ and $f(2)=5$ . The value $6$ is not between $2$ and $5$ . C) False. The IVT does not guarantee uniqueness. While a solution does exist, claiming uniqueness is an incorrect application. D) True. On $[0,2]$ , the function is continuous. Since $f(0)=1$ and $f(2)=5$ , and $4$ is between $1$ and $5$ , the IVT guarantees there is a $c \in (0,2)$ such that $f(c)=4$ .
Let $f(x) = x^3 - 3x + k$ . The Intermediate Value Theorem guarantees that $f(x)$ has a root in the interval $[0, 2]$ for which of the following values of $k$ ?
Answer: $k = -1$ For the Intermediate Value Theorem (IVT) to guarantee a root of $f(x)=0$ on $[0, 2]$ , the function must be continuous on that interval (which it is, as a polynomial), and the value 0 must be between $f(0)$ and $f(2)$ . This occurs if $f(0)$ and $f(2)$ have opposite signs. We evaluate the function at the endpoints: $f(0) = (0)^3 - 3(0) + k = k$ and $f(2) = (2)^3 - 3(2) + k = 8 - 6 + k = 2 + k$ . We need $k(2+k) < 0$ . This inequality holds if and only if $-2 < k < 0$ . Among the given choices, only $k = -1$ is in this interval.
Consider the equation $x^5 + 4x = 3$ . Which statement correctly uses the Intermediate Value Theorem to justify the existence of a solution?
Answer: Let $f(x) = x^5 + 4x - 3$ . Since $f(0)=-3$ and $f(1)=2$ , and $f$ is continuous, a solution must exist in $(0,1)$ . To use the IVT to show a solution exists for $x^5 + 4x = 3$ , we define a new function $f(x) = x^5 + 4x - 3$ and look for its roots. As a polynomial, $f(x)$ is continuous everywhere. We then find an interval $[a, b]$ where $f(a)$ and $f(b)$ have opposite signs. Choice A correctly does this: $f(0) = -3$ and $f(1) = 1+4-3=2$ . Since the signs are opposite, the IVT guarantees a root for $f(x)$ , and thus a solution to the original equation, in the interval $(0, 1)$ . Choices B and D make invalid arguments. Choice C uses a derivative-based argument (related to MVT), not the IVT.
Let $f(x)$ be a continuous function that maps the interval $[0, 1]$ to itself, i.e., for any $x \in [0, 1]$ , $f(x) \in [0, 1]$ . The existence of a fixed point $c \in [0, 1]$ such that $f(c) = c$ is guaranteed by applying the Intermediate Value Theorem to which function on $[0, 1]$ ?
Answer: $g(x) = f(x) - x$ To find a fixed point $f(c)=c$ , we can look for a root of the auxiliary function $g(x) = f(x) - x$ . Since $f(x)$ and $x$ are continuous, $g(x)$ is continuous on $[0, 1]$ . We check the values of $g(x)$ at the endpoints. At $x=0$ , $g(0) = f(0) - 0 = f(0)$ . Since the range of $f$ is $[0, 1]$ , $f(0) \ge 0$ , so $g(0) \ge 0$ . At $x=1$ , $g(1) = f(1) - 1$ . Since the range of $f$ is $[0, 1]$ , $f(1) \le 1$ , so $g(1) \le 0$ . Since $g(x)$ is continuous and its values at the endpoints are on opposite sides of (or on) zero, the IVT guarantees there exists a $c \in [0, 1]$ such that $g(c) = 0$ , which implies $f(c) - c = 0$ , or $f(c) = c$ .
A function $g(x)$ is defined on the interval $[0, 10]$ . It is known that $g(0) = 5$ and $g(10) = 25$ . At $x=4$ , it is observed that $g(4) = 15$ . Based on this information, what must be true about the function $g(x)$ ?
Answer: The function $g(x)$ is not necessarily continuous on the interval $[0, 10]$ . The Intermediate Value Theorem states that if a function is continuous, then it takes on every intermediate value. The converse is not true. A function taking on an intermediate value does not prove that the function is continuous. The function could have a jump discontinuity, for example, but still happen to pass through the value 15. Therefore, we cannot conclude that $g(x)$ must be continuous.
Let $f$ be a continuous function on the closed interval $[a, b]$ . If $f(a) = -2$ and $f(b) = 1$ , which of the following statements about the equation $f(x) = -3$ is correct?
Answer: The IVT is inconclusive regarding solutions, but a solution could exist if $f$ has a minimum value less than or equal to $-3$ . The Intermediate Value Theorem guarantees a solution to $f(x)=N$ if $N$ is a value between $f(a)$ and $f(b)$ . In this case, $f(a)=-2$ and $f(b)=1$ . The value $N=-3$ is not between $-2$ and $1$ . Therefore, the IVT does not guarantee a solution. However, this does not mean a solution cannot exist. The IVT provides a sufficient, not a necessary, condition. The function could dip to a minimum value less than or equal to $-3$ within the interval $(a,b)$ , creating a solution. Thus, the IVT is inconclusive, but a solution might still exist.
Let $f(x) = |x-2| - 1$ . The function is continuous everywhere. We note that $f(0)=1$ and $f(4)=1$ . What can be concluded about the roots of $f(x)$ on the interval $[0,4]$ using the Intermediate Value Theorem?
Answer: The IVT is inconclusive because $f(0)$ and $f(4)$ have the same sign. The Intermediate Value Theorem can only guarantee a root on an interval $[a,b]$ if $f(a)$ and $f(b)$ have opposite signs. In this case, $f(0)=1$ and $f(4)=1$ , which are both positive. Therefore, the IVT is inconclusive; it cannot confirm or deny the existence of a root in $(0,4)$ . Although this particular function does have roots at $x=1$ and $x=3$ , the IVT, based solely on the endpoint values, cannot be used to find them. The differentiability of the function is a condition for the Mean Value Theorem, not the Intermediate Value Theorem, which only requires continuity.

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