Assume that f x , y is differentiable at x 0 , y 0 and let Δ f denote the change in f from its value at x 0 , y 0 to its value at x 0 + Δ x , y 0 + Δ y . (a) Δ f ≈ ______ (b) The limit that guarantees the error in the approximation in part (a) is very small when both Δ x and Δ y are close to 0 is _ _ _ _ _ _ .
Assume that f x , y is differentiable at x 0 , y 0 and let Δ f denote the change in f from its value at x 0 , y 0 to its value at x 0 + Δ x , y 0 + Δ y . (a) Δ f ≈ ______ (b) The limit that guarantees the error in the approximation in part (a) is very small when both Δ x and Δ y are close to 0 is _ _ _ _ _ _ .
Assume that
f
x
,
y
is differentiable at
x
0
,
y
0
and let
Δ
f
denote the change in f from its value at
x
0
,
y
0
to its value at
x
0
+
Δ
x
,
y
0
+
Δ
y
.
(a)
Δ
f
≈
______
(b) The limit that guarantees the error in the approximation in part (a) is very small when both
Δ
x
and
Δ
y
are close to 0 is
_
_
_
_
_
_
.
. The derivative, f', of a function f is plotted below.
At approximately what value of x does f reach a max-
imum, on the range [0, 10]?
N
4
(x) J
2
-4
0
(a) 1
(b) 2.5
(c) 4
(d) 7
(e) 9.5
2
X
8
10
1. The linearization at a = 0 to v8+ &x is A + Bx. Compute A and B.
%3D
A =
2. The linearization at a = 0 to sin(3x) is A + Bx. Compute A and B.
A =
B
3. Find the linearization L(x) of the function g(x) =xf(x²) at x = 2 given the following infor-mation.
f (2) = -1f'(2) = 12f (4) = 6 f'(4) = -2
Answer: L(x) =.
4.
The figure below shows f(x) and its local linearization at x = a, y = 3x – 1. (The local linearization is
shown in blue.)
What is the value of a?
a =
What is the value of f(a)?
f(a) =-
Use the linearization to approximate the value of f(2.5).
f(2.5) =
Is the approximation an under- or overestimate?
(Enter under or over.)
Consider a revenue function defined as
Provide your answer below:
R(x) =
At what value of x is the rate of change of revenue at production level x equal to zero? Give your answer as a decimal to one
decimal place.
Hint: You may wish to make use of the fact that for functions u = u(x), v = v(x)
vu' - uv
(²) ==
where u' and v' denote derivatives of u and v with respect to x.
Content attribution
d
200e0.4x
1 + x
dx
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University Calculus: Early Transcendentals (3rd Edition)
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