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Unformatted text preview: , we know f (2) = 5. The point (2.1, 5.3) is on the tangent
line, so
0.3
5.3 − 5
=
= 3.
Slope tangent =
2.1 − 2
0.1
Thus, f (2) = 3. Since g is the inverse function of f and f (2) = 5, we know f −1 (5) = 2, so g (5) = 2.
Diﬀerentiating, we have
1
1
1
=
=.
g (2) =
f (g (5))
f (2)
3
9. (a) Find dy/dx given that x2 + y 2 − 4x + 7y = 15.
(b) Under what conditions on x and/or y is the tangent line to this curve horizontal? Vertical?
Solution:
(a) By implicit diﬀerentiation, we have:
2x + 2y dy
dy
−4+7
dx
dx
dy
(2y + 7)
dx
dy
dx = 0 = 4 − 2x = 4 − 2x
.
2y + 7 (b) The curve has a horizontal tangent line when dy/dx = 0, which occurs when 4 − 2x = 0 or x = 2. The curve
has a horizontal tangent line at all points where x = 2.
The curve has a vertical tangent line when dy/dx is undeﬁned, which occurs when 2y + 7 = 0 or when
y = −7/2. The curve has a vertical tangent line at all points where y = −7/2.
10. In the following problems ﬁnd the local linearization of f (x) near x = 0 and use this to approximate the value of
a.
(a) f (x) = (1 + x)r , a = (1.2)3/5
Solution: We have f (x) = (1 + x)r , so f (x) = r(1 + x)r−1 . Thus f (0) = r so the local linearization near
x = 0 is
f (x) ≈ f (0) + f (0)x = 1 + rx.
Thus (1 + x)r ≈ 1 + rx for small values of x. Using the linearization with r = 3/5 and x = 0.2, we have
3
1.23/5 = 1 + 0.2 = 1 + 0.12 = 1.12.
5
The actual value is 1.23/5 = 1.117.
(b) f (x) = ekx , a = e0.3
Solution: We have f (x) = ekx , so f (x) = kekx . Thus f (0) = k so the local linearization near x = 0 is
f (x) ≈ f (0) + f (0)x = 1 + kx.
Thus
ekx ≈ 1 + kx for small values of x. Using the linearization with k = 0.3 and x = 1, we have
e0.3 ≈ 1 + 0.3 = 1.3
The actual value is e0.3 = 1.350. √
√
(c) f (x) = b2 + x, a = 26
Solution: We have f (x) = (b2 + x)1/2 , so f (x) = (1/2)(b2 + x)−1/2 . Thus f (0) = 1/(2b) so the local
Math 115 / Exam 2 (March 20, 2012)
page 4
linearization near x = 0 is
1
f (x) ≈ f (0) + f (0)x = b + x.
2b
3. [12 points] The following questions relate to the implicit function
Thus
1
Math 115 / Exam 2 (March 20, 2012)
page 4
2x
b2y+ + ≈x = 4xy 2 . for small values of x.
4 b+ x
2b
Using the linearizationdy
with b = 5 and x = 1, we have 3. [12. points] The following questions relate to the implicit function
.
a [4 points] Compute
dx
√
1
26 ≈ 5 + 1 = 5.1.
2
y + 4x = 4xy 210
.
Solution: Diﬀerentiating the equation with respect to x, we have Math 115 / Exam 2 (March 20, 2012)
√ dy
dy
2y
+ 4 = 4y 2 + 8xy .
.
a. [4 points] Compute
dx
dx
dx
11. 3Let y 2points] 4The following questions relate to the implicit function
. [12 + 4x = xy 2
Solution: Diﬀerentiating dy equation the respect becomes
the
Gathering terms involving dx to one side,with equationto x, we have
dy
(a) Compute dx .
y 2 + 4x = 4xy 2 .
dy this curve 2 (1/3, 2)
(b) Find the equation for the tangent line to + 4 = 4y at 2 xy dy .
dy
dy
+ 8− 4
22y dx 8
dx
dy of all y dx −atxy dx = 4ytangent line to this curve is vertical.
(c) a. [4 points] Compute
Find the x and y coordinates
points...
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This note was uploaded on 11/19/2012 for the course ECON CALC 1 taught by Professor Fisherselin during the Fall '11 term at NYU.
 Fall '11
 fisherselin

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