Find all points of discontinuity of f, where f is defined by 17.t – t² 0119.Se-t 0 2

Answered: St – t² 0 1 f (t) = = let 0 2 f (t) =

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter1: Fundamental Concepts Of Algebra

Section1.2: Exponents And Radicals

Problem 28E

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Question

Find all points of discontinuity of f, where f is defined by

17.
t – t² 0<t< 1
-
f(t) =
三
t>1
19.
Se-t 0<t<2
f (t) = { 0
t > 2

Expert Solution

Step 1: Definition

A function is continuous at a point $t = a$ if it satisfies all the given conditions.

$f (a)$ exist
$\lim_{t \to a} f (t)$ exist. That is, $\lim_{t \to a^{-}} f (t) = \lim_{t \to a^{+}} f (t)$
The limit is equal to the value of the function at that point. That is, $\lim_{t \to a} f (t) = f (a)$ .

Step 2

(17)

The function is $f (t) = \{\begin{cases} t - t^{2}, 0 \leq t < 1 \\ 0, t \geq 1 \end{cases}$ .

The function is a polynomial function in the intervals $[0, 1)$ and constant in $(1, \infty)$ . Thus, the function is continuous there.

Now, check the continuity of the function at the point $t = 1$ .

The value of $f (1) = 0$ .

Find the limit of the function as $t \to 1$ .

$\begin{array}{rcl} \lim_{t \to 1^{-}} f (t) & = & \lim_{t \to 1^{-}} t - t^{2} \\ = & 1 - 1^{2} \\ = & 0 \\ \lim_{t \to 1^{+}} f (t) & = & \lim_{t \to 1^{+}} 0 \\ = & 0 \\ \lim_{t \to 1^{-}} f (t) & = & \lim_{t \to 1^{+}} f (t) \end{array}$

Therefore, $\lim_{t \to 1} f (t) = f (1) = 0$ .

Thus, the function is continuous at $t = 1$ .

That is, the function is continuous at all points in the domain.

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