findI+ 226. g(1) :I- 29(-2). g(2). g(0). g(a), g(a² – 2). g(a + 1)27. k(x) = -x - 2r + 3;k(0), K(2), k( –2), k( v2), k(a +2), k( –x), k(x²)

Answered: find I+ 2 26. g(1) : I- 2 9(-2). g(2).…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

find
I+ 2
26. g(1) :
I- 2
9(-2). g(2). g(0). g(a), g(a² – 2). g(a + 1)
27. k(x) = -x - 2r + 3;
k(0), K(2), k( –2), k( v2), k(a +
2), k( –x), k(x²)

Expert Solution

Step 1

We’ll answer the 26th question since we answer only one question at a time. Please submit a new question specifying the one you’d like answered.

Given function,

$g (t) = \frac{t + 2}{t - 2} . . . (1)$

Step 2

Put $t = - 2$ in g(t)

$g (- 2) = \frac{- 2 + 2}{- 2 - 2} g (- 2) = \frac{0}{- 4} g (- 2) = 0$

Put $t = 2$ in g(t)

$g (2) = \frac{2 + 2}{2 - 2} g (2) = \frac{4}{0} g (2) = \infty$

Put $t = 0$ in g(t)

$g (0) = \frac{0 + 2}{0 - 2} g (0) = \frac{2}{- 2} g (0) = - 1$

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find I+ 2 26. g(1) : I- 2 9(-2). g(2). g(0). g(a), g(a² – 2). g(a + 1) 27. k(x) = -x - 2r + 3; k(0), K(2), k( –2), k( v2), k(a + 2), k( –x), k(x²)