Answer
$f(t)=\sqrt {t+1}, a=1$
Work Step by Step
Let $f(t)=\sqrt {t+1}$, we have:
$$f'(1)=\lim_{t\to 1}\frac{\sqrt {t+1}-\sqrt {1+1}}{t-1}=\lim_{t\to 1}\frac{\sqrt {t+1}-\sqrt {2}}{t-1}$$
Thus we have $f(t)=\sqrt {t+1}, a=1$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 13 - Section 13.3 - Tangent Lines and Derivatives - 13.3 Exercises - Page 922: 35
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
