Thomas Calculus 13th [Solutions]

198 Chapter 3 Derivatives
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f ) The linearization of any differentiable function u( x) at x a is L( x ) u( a) u ( a)( x a)
b 0 b 1 ( x a),
where b 0 and b 1 are the coefficients of the constant and linear terms of the quadratic approximation. Thus,
the linearization for f ( x ) at x 0 is 1 x ; the linearization for g( x) at x 1is 1 ( x 1) or 2 x ; and the
linearization for h( x) at x 0 is 1 x .
2
66. E( x) f ( x) g( x) E( x ) f ( x) m( x a) c. Then E( a ) 0 f ( a) m( a a)
c 0 c f ( a).
( )
Next we calculate : lim E x
( ) ( )
m
0 lim f x m x a c
( ) ( )
0 lim f x f a m 0 (since c f ( a))
x a
x a
x a
x a
x a
x a
f ( a)
m 0 m f ( a ). Therefore, g( x) m( x a)
c f ( a)( x a) f ( a ) is the linear approximation,
as claimed.
x
x
0 0
67. (a) f ( x) 2 f ( x) 2 ln 2; L( x) (2 ln 2) x 2 x ln 2 1 0.69x
1
(b)
68. (a) f ( x) log 3 x f ( x ) 1
ln 3 , and f (3) ln 3
L( x) 1 ( x 3) ln3 x 1 1
x
ln 3 3ln 3 ln3 3ln 3 ln 3
(b)
69 74. Example CAS commands:
Maple:
with(plots):
a : 1: f: x -> x^3 x^2 2*x;
plot(f(x), x 1..2);
diff (f(x), x);
fp : unapply ( , x);
L: x ->f(a) fp(a)*(x a);
plot({f(x), L(x)}, x 1..2);
err: x -> abs(f(x) L(x));
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