Continuity on an Interval, Types of Discontinuities, and the IVT and EVT
Marcus merges onto the highway from a stop sign: his speedometer reads 0 mph, and ten seconds later it reads 65 mph. Did it read exactly 45 mph at some moment? You never saw the dial, yet you are sure it did. Continuity is the reason, and in this section you will learn to turn that feeling into a proof that earns full credit on an AP free-response question.
Continuity on an Interval
You already know continuity at a single point. Let's restate it as a three-part checklist, because every problem in this section is built on it.
Now stretch this over a whole interval. At an endpoint of a closed interval you can only approach from one side, so the condition uses just that one-sided limit.
To justify continuity quickly, use the toolbox: polynomials, $\sin x$, $\cos x$, $e^x$ and $|x|$ are continuous everywhere. $\ln x$ is continuous on $(0,\infty)$ and $\sqrt{x}$ on $[0,\infty)$. A rational function is continuous everywhere its denominator is not zero. Sums, differences, products, quotients (denominator $\neq 0$) and compositions of continuous functions are continuous. For a piecewise function, each piece is fine on its own, so only the breakpoints need checking.
Study the one-sided continuity of the following function at the endpoints of the interval $[0, 4]$: $$f(x) = \begin{cases} \sqrt{x} & \text{if } 0 \le x < 1 \\ x^2 - x +…
Study the one-sided continuity of the following function at the endpoints of the interval $[0, 4]$:
$$f(x) = \begin{cases} \sqrt{x} & \text{if } 0 \le x < 1 \\ x^2 - x + 1 & \text{if } 1 \le x \le 4 \end{cases}$$Indicate whether the function is continuous at $x=1$ by comparing its one-sided limits.
Types of Discontinuities
A discontinuity at $x=c$ means one of the three conditions fails. How it fails gives the type.
Each type has its own signature. A removable discontinuity can be repaired by redefining $f(c)$ as the limit. A jump cannot be repaired, because the two sides disagree. An infinite discontinuity cannot be repaired either.
For rational functions, factor and look at what cancels. A factor that cancels from numerator and denominator leaves a hole (removable). A factor of the denominator that survives, while the numerator is not zero there, leaves a vertical asymptote (infinite).
Observe the graph of the function $f(x)$ represented below and identify the points of discontinuity. For each of them, indicate what type it is (removable, finite jump…
Observe the graph of the function $f(x)$ represented below and identify the points of discontinuity. For each of them, indicate what type it is (removable, finite jump or infinite jump).
Consider the function $f(x) = \frac{x^2 - 4x + 4}{x - 2}$. 1. Prove that the function is not continuous at $x = 2$. 2. What value should $f(2)$ take so that the…
Consider the function $f(x) = \frac{x^2 - 4x + 4}{x - 2}$.
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Prove that the function is not continuous at $x = 2$.
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What value should $f(2)$ take so that the discontinuity is removable and the function becomes continuous at that point?
Given the rational function $f(x) = \frac{x-3}{x^2-9}$, study its continuity at the points $x = 3$ and $x = -3$. If there are any discontinuities, classify them by…
Given the rational function $f(x) = \frac{x-3}{x^2-9}$, study its continuity at the points $x = 3$ and $x = -3$. If there are any discontinuities, classify them by calculating their one-sided limits and justifying your answer.
Finding Parameters That Make a Function Continuous
A very common AP-style task: a piecewise function has unknown constants, and you must choose them so that $f$ is continuous. The idea is simple. At every breakpoint $c$, force the left limit, the right limit, and $f(c)$ to be equal.
Some parameter problems need harder limits. Here are three tools.
Calculate the value of the parameter $k$ so that the following piecewise function is continuous over the set of all real numbers: $$f(x) = \begin{cases} x^2 - 1 &…
Calculate the value of the parameter $k$ so that the following piecewise function is continuous over the set of all real numbers:
$$f(x) = \begin{cases} x^2 - 1 & \text{if } x < 2 \\ 2x + k & \text{if } x \ge 2 \end{cases}$$Consider the piecewise function $f(x)$ defined as: $$f(x) = \begin{cases} \frac{x^2 - a}{x - 2} & \text{if } x < 2 \\ 2x + b & \text{if } 2 \le x < 4 \\ \ln(x - 3) + k &…
Consider the piecewise function $f(x)$ defined as:
$$f(x) = \begin{cases} \frac{x^2 - a}{x - 2} & \text{if } x < 2 \\ 2x + b & \text{if } 2 \le x < 4 \\ \ln(x - 3) + k & \text{if } x \ge 4 \end{cases}$$Calculate the values of $a$, $b$ and $k$ so that the function is continuous throughout its domain $\mathbb{R}$. Once found, explain whether the function satisfies the hypotheses of Weierstrass's Theorem in the interval $[1, 5]$.
Given the piecewise function $f(x)$ defined by the following expression, determine the values of the real parameters $a$ and $b$ so that the function is continuous over…
Given the piecewise function $f(x)$ defined by the following expression, determine the values of the real parameters $a$ and $b$ so that the function is continuous over its entire domain. In the case that $b=1$, classify the type of discontinuity that the function presents at $x=0$ depending on the value of $a$.
$$f(x) = \begin{cases} \frac{e^{ax} - \cos(x)}{x} & \text{if } x < 0 \\ b & \text{if } x = 0 \\ \left(1 + \sin(x)\right)^{\frac{2}{x}} & \text{if } x > 0 \end{cases}$$Classifying Harder Functions
Now let's handle functions with several suspicious points at once: rational functions with a parameter, absolute values, and functions defined by a limit.
For absolute values, split into cases using the sign of the inside expression. Since $|u|=u$ when $u\ge0$ and $|u|=-u$ when $u<0$, find where the inside is zero first.
Sometimes a function is defined as a limit in a variable $n$ (not in $x$). Treat $x$ as a fixed number, find the limit for each case of $x$, and you get an ordinary piecewise function. You will need: $x^n\to0$ if $|x|<1$, $x^n\to\infty$ if $x>1$, and $1^n=1$.
Study the continuity of the following function and classify all its points of discontinuity, justifying your answer: $$f(x) = \frac{x^2 - x - 6}{x^2 + x - 12}$$ Indicate…
Study the continuity of the following function and classify all its points of discontinuity, justifying your answer:
$$f(x) = \frac{x^2 - x - 6}{x^2 + x - 12}$$Indicate if any of the discontinuities is removable and, if so, redefine the function at that point so that it becomes continuous.
Consider the family of functions dependent on the parameter $k \in \mathbb{R}$: $$f(x) = \frac{x^2 - (k+1)x + k}{x^3 - 3x^2 + 2x}$$ Analyze and classify all the…
Consider the family of functions dependent on the parameter $k \in \mathbb{R}$:
$$f(x) = \frac{x^2 - (k+1)x + k}{x^3 - 3x^2 + 2x}$$Analyze and classify all the discontinuities of $f(x)$ depending on the possible values of $k$. Indicate specifically for which values of $k$ any of the discontinuities is removable and redefine the function at that point so that it is continuous.
Study the continuity of the following function in its domain of definition: $$f(x) = \lim_{n \to \infty} \frac{x^n - 1}{x^n + 1}$$ Determine the points where the…
Study the continuity of the following function in its domain of definition:
$$f(x) = \lim_{n \to \infty} \frac{x^n - 1}{x^n + 1}$$Determine the points where the function presents discontinuities and classify them. Investigate whether the function satisfies the hypotheses of Bolzano's Theorem in the interval $[0.5, 2]$ to ensure the existence of a root.
The Intermediate Value Theorem
Back to the on-ramp. Marcus's speed is a continuous function of time: it cannot teleport from 0 to 65 mph. A continuous graph is one unbroken curve. To get from the height $f(a)$ to the height $f(b)$, it must cross every level in between.
An important special case: if $f$ is continuous on $[a,b]$ and $f(a)$, $f(b)$ have opposite signs, then $N=0$ lies between them and $f$ has a root in $(a,b)$. This is the standard way to prove a root exists without solving the equation.
A hiker starts a route at a refuge at an altitude of 1,200 feet ($t=0$). After 4 hours of walking, they are at a viewpoint at an altitude of 1,800 feet. Assuming that…
A hiker starts a route at a refuge at an altitude of 1,200 feet ($t=0$). After 4 hours of walking, they are at a viewpoint at an altitude of 1,800 feet. Assuming that the function describing their altitude with respect to time, $A(t)$, is continuous:
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Explain, using the Intermediate Value Theorem (IVT), whether the hiker has passed the 1,500-foot mark at any point.
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Can we guarantee that they have passed the 2,000-foot mark?
Verify whether the function $f(x) = x^3 - 3x + 1$ satisfies the hypotheses of Bolzano's Theorem in the interval $[0, 1]$. Can you guarantee that there is at least one…
Verify whether the function $f(x) = x^3 - 3x + 1$ satisfies the hypotheses of Bolzano's Theorem in the interval $[0, 1]$. Can you guarantee that there is at least one root of the equation $x^3 - 3x + 1 = 0$ in this interval? Justify your answer.
Given the function $f(x) = x^3 + x - 5$, justify whether there is any value $c \in [1, 2]$ such that $f(c) = -1$. Which theorem guarantees this result? State this…
Given the function $f(x) = x^3 + x - 5$, justify whether there is any value $c \in [1, 2]$ such that $f(c) = -1$. Which theorem guarantees this result? State this theorem and verify that its hypotheses are met.
Roots and Intersections
Some problems call the sign-change version of the IVT Bolzano's Theorem. It is the tool for proving that an equation has a solution. The trick for equations like $f(x)=g(x)$ is to move everything to one side.
When the two functions are real-world models (heights of two drones, prices of two plans), the same idea shows they are equal at some moment: let $d(t)$ be the difference and look for a sign change.
Prove, using Bolzano's Theorem, that the graphs of the functions $f(x) = e^x - 2$ and $g(x) = \cos(x)$ intersect at at least one point within the interval $[0, \pi]$.…
Prove, using Bolzano's Theorem, that the graphs of the functions $f(x) = e^x - 2$ and $g(x) = \cos(x)$ intersect at at least one point within the interval $[0, \pi]$. Explain in detail the steps and hypotheses necessary to apply the theorem.
A team of engineers is testing two new models of surveillance drones, $D_1$ and $D_2$. The height (in feet) of drone $D_1$ with respect to the ground is given by the…
A team of engineers is testing two new models of surveillance drones, $D_1$ and $D_2$. The height (in feet) of drone $D_1$ with respect to the ground is given by the function $h_1(t) = t \cdot \ln(t+1)$, while the height of drone $D_2$ follows the function $h_2(t) = 10 \cdot e^{-t} + 5$, where $t$ is the time in minutes since take-off ($t \geq 0$).
Use Bolzano's Theorem to prove that there is an instant $t_0 \in [2, 5]$ at which both drones are at exactly the same height. Justify in detail whether you could ensure, using Weierstrass's Theorem, that the difference in height between both drones reaches a maximum value in that time interval.
Consider a continuous function $f:[0, 1] \to [0, 1]$. That is, the domain is the interval $[0, 1]$ and all its images are also contained in $[0, 1]$. Prove, using…
Consider a continuous function $f:[0, 1] \to [0, 1]$. That is, the domain is the interval $[0, 1]$ and all its images are also contained in $[0, 1]$.
Prove, using Bolzano's Theorem, that there is at least one point $c \in [0, 1]$ such that $f(c) = c$ (a fixed point of the function). Hint: Consider the auxiliary function $g(x) = f(x) - x$. Explain graphically what this result means in terms of the intersection of the graph of $f(x)$ with the angle bisector of the first quadrant.
The Extreme Value Theorem
The IVT is about values between. The next theorem is about the highest and lowest values: a continuous function on a closed interval cannot keep climbing forever or sneak away from its minimum.
Some problems call this result Weierstrass's Theorem.
Why does this make sense? The graph is one unbroken piece with no gaps and no escape to infinity, and both endpoints are included. So the set of output values is itself a closed interval $[m,M]$, which contains its own lowest and highest values.
Explain whether the function $f(x) = \frac{1}{x}$ necessarily reaches an absolute maximum and minimum in the following intervals based on the Extreme Value Theorem…
Explain whether the function $f(x) = \frac{1}{x}$ necessarily reaches an absolute maximum and minimum in the following intervals based on the Extreme Value Theorem (EVT):
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In the closed interval $[1, 5]$.
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In the open interval $(0, 1)$.
Given the function $f(x) = \frac{k \cdot \sin(x)}{x}$ for $x \neq 0$: a) Find the value of $k$ knowing that $\lim_{x \to 0} f(x) = 5$. b) With that value of $k$, define…
Given the function $f(x) = \frac{k \cdot \sin(x)}{x}$ for $x \neq 0$:
- a) Find the value of $k$ knowing that $\lim_{x \to 0} f(x) = 5$.
- b) With that value of $k$, define $f(0)$ so that the function is continuous at $x = 0$.
- c) Justify whether the resulting function is continuous in the interval $[0, 2\pi]$. Can we guarantee that it reaches an absolute maximum value in this interval?
Let $f(x) = |x^2 - 4| \cdot e^{-|x|}$ be the function. 1. Express the function without moduli. 2. Study its continuity at the points where the arguments of the moduli…
Let $f(x) = |x^2 - 4| \cdot e^{-|x|}$ be the function.
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Express the function without moduli.
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Study its continuity at the points where the arguments of the moduli are zero.
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Using Weierstrass's Theorem, prove that the function reaches an absolute maximum in the closed interval $[-3, 3]$.
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Does the function have any horizontal asymptote? Relate this fact to the long-term behavior of the function.
Why the Hypotheses Matter
Both theorems stop working when a hypothesis fails. A single discontinuity inside the interval is enough to break the IVT, because the graph can leap over the level $N$.
A graphic designer is plotting the trajectory of an object in an animation. The height $h$ in feet with respect to time $t$ (in seconds) is given by: $$h(t) = \frac{|t -…
A graphic designer is plotting the trajectory of an object in an animation. The height $h$ in feet with respect to time $t$ (in seconds) is given by:
$$h(t) = \frac{|t - 3|}{t^2 - 9}$$- a) State the domain of the trajectory.
- b) Study the continuity at $t = 3$. What type of discontinuity does it present? Interpret the result in the context of the animation (is there a jump, does the object disappear, or does it go to infinity?).
- c) Is it possible for the object to reach a height of $0$ feet at any instant in the interval $[0, 2]$? Justify your answer using Bolzano's Theorem.
It is known that a function $f(x)$ is continuous in the closed interval $[0, 4]$. If $f(0) = 1$ and $f(4) = 1$, can it be guaranteed that there is a point $c \in (0, 4)$…
It is known that a function $f(x)$ is continuous in the closed interval $[0, 4]$. If $f(0) = 1$ and $f(4) = 1$, can it be guaranteed that there is a point $c \in (0, 4)$ such that $f(c) = 0$? Give a graphical example of a function that meets the conditions but has no roots in that interval, and explain why it does not contradict Bolzano's Theorem.
Given the function $f(x) = \frac{1}{1 + 2^{1/x}}$: 1. Find the domain of the function. 2. Calculate the one-sided limits at the point where the function is not defined.…
Given the function $f(x) = \frac{1}{1 + 2^{1/x}}$:
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Find the domain of the function.
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Calculate the one-sided limits at the point where the function is not defined.
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Classify the discontinuity found.
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Is it possible to apply Bolzano's Theorem in the interval $[-1, 1]$ to prove that there is a value where $f(x) = 0.5$? Explain whether the lack of continuity at a point invalidates the application of the theorem even if the values at the endpoints suggest that such a point should exist.