Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises - Page 124: 22

Answer

The function is discontinuous because $f(3) = 6 \ne 7$.
1513721324

Work Step by Step

For this function to be continuous by definition: $\lim\limits_{x \to 3} f(x) = f(3)$ $ \lim\limits_{x \to 3} \frac{2x^{2}-5x-3}{x-3} = $ $ \lim\limits_{x \to 3} \frac{(2x+1)(x-3)}{x-3} =$ Cancel out $(x-3)$ $ \lim\limits_{x \to 3} (2x+1) =$ $ \lim\limits_{x \to 3} (2(3)+1) = 7$ The value of $f(3) = 6$, but we discovered that $ \lim\limits_{x \to 3} (2(3)+1) = 7$ so the function is discontinuous. To draw the graph we draw the function of $f(x) = (2x+1)$ and $(3,6)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.