Question

use the graph of the function t shown to estimate the indicated quantities to the nearest integer. complete parts a through e. a. t(1) = 1 b. the value does not exist. e. is t continuous at x=1?
no
yes

Explanation:

Part e: Is \( f \) continuous at \( x = 1 \)?

To determine if a function \( f \) is continuous at a point \( x = a \), three conditions must be met:

1. \( f(a) \) is defined.
2. \( \lim_{x \to a} f(x) \) exists (i.e., the left - hand limit \( \lim_{x \to a^{-}} f(x) \) and the right - hand limit \( \lim_{x \to a^{+}} f(x) \) are equal).
3. \( \lim_{x \to a} f(x)=f(a) \).

Step 1: Check if \( f(1) \) is defined

From part (a), we know that \( f(1) = 1 \), so \( f(1) \) is defined.

Step 2: Check the left - hand limit as \( x \to 1^{-} \)

As \( x \) approaches \( 1 \) from the left (values of \( x \) less than \( 1 \)), we look at the graph of the function. From the graph, as \( x \) gets closer to \( 1 \) from the left, the \( y \) - values approach \( 1 \). So, \( \lim_{x \to 1^{-}} f(x)=1 \).

Step 3: Check the right - hand limit as \( x \to 1^{+} \)

As \( x \) approaches \( 1 \) from the right (values of \( x \) greater than \( 1 \)), we look at the graph of the function. From the graph, as \( x \) gets closer to \( 1 \) from the right, the \( y \) - values also approach \( 1 \). So, \( \lim_{x \to 1^{+}} f(x)=1 \).

Since \( \lim_{x \to 1^{-}} f(x)=\lim_{x \to 1^{+}} f(x) = 1 \), the limit \( \lim_{x \to 1} f(x) \) exists and is equal to \( 1 \). And we already know that \( f(1)=1 \), so \( \lim_{x \to 1} f(x)=f(1) \).

All three conditions for continuity are satisfied.

Answer:

Yes