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Tutorial Exercise Find the linear approximation of the function g(x) = 1 + x at a = 0. Use it to approximate the numbers 0.95 and 1.1. Use technology to graph g and the tangent line in the same viewing window. Step 1 Recall that the linearization of g(x) for x = a is given by L(x) = g(a) + g'(a)(x - a). The first step is to find g'(x) so that it can be used to find g'(a). √√1 + x = (1 + x) , we have 1/9 1 For g(x) g'(x) = = Therefore, at a = 0 we have. g'(0)=1/9✔✓ 1/9 Step 2 Now, at a = 0, we have g(0) = Therefore, at a = 0, we have g(x) = 1 + x = g(0) + g'(0)(x - 0) (1 + x) 0.95 = ≈ 1 + We will now use this linear approximation to approximate 0.95. We first need to rewrite 0.95 in the form 1 + x, where x is a numerical value. -8 /9

Tutorial Exercise Find the linear approximation of the function g(x) = 1 + x at a = 0. Use it to approximate the numbers 0.95 and 1.1. Use technology to graph g and the tangent line in the same viewing window. Step 1 Recall that the linearization of g(x) for x = a is given by L(x) = g(a) + g'(a)(x - a). The first step is to find g'(x) so that it can be used to find g'(a). √√1 + x = (1 + x) , we have 1/9 1 For g(x) g'(x) = = Therefore, at a = 0 we have. g'(0)=1/9✔✓ 1/9 Step 2 Now, at a = 0, we have g(0) = Therefore, at a = 0, we have g(x) = 1 + x = g(0) + g'(0)(x - 0) (1 + x) 0.95 = ≈ 1 + We will now use this linear approximation to approximate 0.95. We first need to rewrite 0.95 in the form 1 + x, where x is a numerical value. -8 /9

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter5: A Survey Of Other Common Functions
Section5.4: Combining And Decomposing Functions
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Tutorial Exercise Find the linear approximation of the function g(x) V1 + x at a = 0. Use it to approximate the numbers 0.95 and 1.1. Use technology to graph g and the tangent line in the same viewing window. Step 1 Recall that the linearization of g(x) for x = a is given by L(x) = g(a) + g'(a)(x − a). The first step is to find g'(x) so that it can be used to find g'(a). √√1 + x = (1 + x) ¹/⁹, we have 1/9 1 For g(x) = g'(x): g'(0) Therefore, at a = 0 we have 1/9 = 9 g(x) 9 = (1 + x) Step 2 Now, at a = 0, we have g(0) = = Therefore, at a = 0, we have 1/9 0.95 = 1 + -8 9 1 + x g(0) + g'(0)(x −0) = -8/9 We will now use this linear approximation to approximate 0.95. We first need to rewrite 0.95 in the form 1 + x, where x is a numerical value.
Transcribed Image Text:Tutorial Exercise Find the linear approximation of the function g(x) V1 + x at a = 0. Use it to approximate the numbers 0.95 and 1.1. Use technology to graph g and the tangent line in the same viewing window. Step 1 Recall that the linearization of g(x) for x = a is given by L(x) = g(a) + g'(a)(x − a). The first step is to find g'(x) so that it can be used to find g'(a). √√1 + x = (1 + x) ¹/⁹, we have 1/9 1 For g(x) = g'(x): g'(0) Therefore, at a = 0 we have 1/9 = 9 g(x) 9 = (1 + x) Step 2 Now, at a = 0, we have g(0) = = Therefore, at a = 0, we have 1/9 0.95 = 1 + -8 9 1 + x g(0) + g'(0)(x −0) = -8/9 We will now use this linear approximation to approximate 0.95. We first need to rewrite 0.95 in the form 1 + x, where x is a numerical value.
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