Continuity on an interval. Classification of discontinuities. Bolzano's and Weierstrass's theorems

Study the one-sided continuity of the following function at the endpoints of the interval $[0, 4]$:

State whether the function is continuous at $x=1$ by comparing its left-hand and right-hand limits.

Calculate the value of the parameter $k$ so that the following piecewise function is continuous on the set of all real numbers:

Given the function $f(x) = x^3 + x - 5$, justify whether there exists a value $c \in [1, 2]$ such that $f(c) = -1$. What theorem guarantees this result? State the theorem and verify that its hypotheses are satisfied.

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 discontinuities, classify them, giving reasons by calculating their one-sided limits.

Check whether the function $f(x) = x^3 - 3x + 1$ satisfies the hypotheses of Bolzano's Theorem on the interval $[0, 1]$. Can you guarantee that there is at least one root of the equation $x^3 - 3x + 1 = 0$ in that interval? Justify your answer.

A team of engineers is testing two new models of surveillance drones, $D_1$ and $D_2$. The altitude (in metres) of drone $D_1$ above the ground is given by the function $h_1(t) = t \cdot \ln(t+1)$, while the altitude of drone $D_2$ follows the function $h_2(t) = 10 \cdot e^{-t} + 5$, where $t$ is the time in minutes since takeoff ($t \geq 0$).

Use Bolzano's Theorem to prove that there is a time $t_0 \in [2, 5]$ at which both drones are at exactly the same altitude. Justify in detail whether you could guarantee, using Weierstrass's Theorem, that the difference in altitude 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 Bolzano's Theorem, that there exists 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 line $y=x$ (the bisector of the first quadrant).

Explain whether the function $f(x) = \frac{1}{x}$ must reach an absolute maximum and minimum on the following intervals, based on Weierstrass's Theorem:

1. On the closed interval $[1, 5]$.

2. On the open interval $(0, 1)$.

Let $f(x)$ be the piecewise function defined as:

The figure shows the three defining pieces (they change at $x=2$ and at $x=4$) and the interval $[1, 5]$.

1. a) Calculate the values of $a$, $b$ and $k$ so that the function is continuous on its entire domain $\mathbb{R}$.

1. b) Once you have found them, explain whether the function satisfies the hypotheses of the Weierstrass Theorem on the interval $[1, 5]$.

Let the function be $f(x) = |x^2 - 4| \cdot e^{-|x|}$. The figure shows its graph as a visual aid.

1. a) Express the function without absolute values, that is, defined piecewise.

1. b) Study its continuity at the points where the arguments of the absolute values vanish.

1. c) Using the Weierstrass Theorem, prove that the function attains an absolute maximum on the closed interval $[-3, 3]$.

1. d) Does the function have any horizontal asymptote? Relate this fact to the long-term behavior of the function.

Reasonably prove that the equation $\sqrt{x^4 + 1} = 3x^2 \cdot \cos(x)$ has at least two real solutions in the interval $[-\pi, \pi]$. To do this, define an appropriate auxiliary function and apply the necessary continuity theorems. Is it possible to state that an absolute maximum value exists for the function $g(x) = \frac{1}{\sqrt{x^4+1} - 3x^2\cos(x)}$ in the interval $[0, \pi/4]$? Justify your answer based on the Weierstrass Theorem.

Given the function $f(x) = \frac{1}{1 + 2^{1/x}}$:

1. Find the domain of the function.

2. Calculate the left-hand and right-hand limits at the point where the function is not defined.

3. Classify the discontinuity found.

4. Is it possible to apply Bolzano's Theorem on the interval $[-1, 1]$ to prove that there exists 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.