EXAMPLE 1 Find the linearization of the function f(x) = Vx + 7 at a = 2 and use it to approximate the numbers 8.95 and 9.05. Are these approximations overestimates or underestimates? SOLUTION The derivative of f(x) = (x + 7)/2 is f'(x) = and so we have (2) = O and f'(2) - . Putting these values into this equation, we see that the linearization is L(x) = f(2) + l'(2)(x – 2) - + (x - 2) -7 Video Example) The corresponding linear approximation is Vx+7 = (when x is near 2). In particular, we have V8.95 =+ (round to four decimal places) and V9.05 = (round to four decimal places). The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 2. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for v8.95 and v9.05, but the linear approximation gives an approximation over an entire interval.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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Question
y
EXAMPLE 1
Find the linearization of the function f(x)
Vx + 7 at a = 2 and use it to approximate the numbers
8.95 and
9.05. Are these
=
approximations overestimates or underestimates?
3
SOLUTION
The derivative of f(x) = (x + 7)1/2 is
f'(x) =
and so we have f(2)
and f'(2)
Putting these values into this equation, we see that the linearization is
%3D
L(x) = f(2) + f '(2)(x – 2) =
+
(х — 2)
-
-7
2
Video Example )
The corresponding linear approximation is
Vx + 7 =
+
(when x is near 2).
In particular, we have
8
8.95 =
+
(round to four decimal places)
3
6.
and
8
V9.05 z+
6.
(round to four decimal places).
%D
The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given
function when x is near 2. We also see that our approximations are overestimates because the tangent line lies above the curve.
Of course, a calculator could give us approximations for v 8.95 and V9.05, but the linear approximation gives an approximation over an entire
interval.
Transcribed Image Text:y EXAMPLE 1 Find the linearization of the function f(x) Vx + 7 at a = 2 and use it to approximate the numbers 8.95 and 9.05. Are these = approximations overestimates or underestimates? 3 SOLUTION The derivative of f(x) = (x + 7)1/2 is f'(x) = and so we have f(2) and f'(2) Putting these values into this equation, we see that the linearization is %3D L(x) = f(2) + f '(2)(x – 2) = + (х — 2) - -7 2 Video Example ) The corresponding linear approximation is Vx + 7 = + (when x is near 2). In particular, we have 8 8.95 = + (round to four decimal places) 3 6. and 8 V9.05 z+ 6. (round to four decimal places). %D The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 2. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for v 8.95 and V9.05, but the linear approximation gives an approximation over an entire interval.
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