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AMS 151 (Fall, 2009)
Joe Mitchell
Applied Calculus I
Quiz # 4, Thursday, October 22 – Solution Notes
1. Find the equation of the tangent line to the curve
y
= 2
√
x
+ 1 (
x
≥
0) at the point (4,5).
Since the slope is given by the derivative function,
y
′
(
x
) = 2
·
(1
/
2)
x
−
1
/
2
=
1
√
x
, we know that the slope of the
tangent line at (4,5) is
y
′
(4) = 1
/
2. Thus, the equation of the tangent line at (4,5) has the form
y
= (1
/
2)
x
+
b
, and
we can ±nd
b
using the fact that the line must pass through the point (4,5): 5 = (1
/
2)
·
4 +
b
, implying that
b
= 3.
Thus, the equation of the tangent line at (4,5) is
y
= (1
/
2)
x
+ 3.
2. If
h
(
b
) =
xb
3
+
b
2

x
2
b
, ±nd
h
′
(1). Also ±nd
h
′′
(
x
).
h
′
(
b
) =
x
·
3
b
2
+ 2
b

x
2
·
(

1)
b
−
2
= 3
xb
2
+ 2
b
+
x
2
b
2
. Taking the second derivative, we get
h
′′
(
b
) = 6
xb
+ 2 +
x
2
·
(

2)
b
−
3
= 6
xb
+ 2

2
x
2
b
3
.
Thus,
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This note was uploaded on 02/07/2010 for the course AMS 151 taught by Professor Zhang during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Zhang

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