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Unformatted text preview: Math 2123.2 — Fall 2011 — Sample Exam 2 Solution To receive credit, you must show your work. 1. [15 points] Complete the values in the table and use them to guess 2 6—1
MEL")— m—>O [1: to as much accuracy as the table allows.
(2+x)x~1 x as The table allows us to guess
0.01 0.70057 m
lim Old—1 z 0.69
0.001 0.69388 9590 1;
0 ? or even
—0.001 0.69240 ac _ 1
0.6924 < lim ————<2 + m) < 0.6939.
—0.01 0.68577 :r—>0 r 2. [12 points] Find an equation of the tangent line to the curve y = x5 — 5232 at the point
Where :6 = —1. When a: = —1, y‘w:_1 = (—1)5 — 5(—1)2 = —6 and so the point is (—1,—6). The
derivative of y is
y’ = 5:54 — 5(2m) = 5.734 — 10.6 and so the slope of the tangent line is
y’lxﬂ = 5(—1)4 —10(—1) = 15.
By the pointslope form, an equation of the tangent line is y——6=15(m——1)
y+6=15(a:+1). This may also be written as
y : 15:13 + 9. 5. [15 points] Use appropriate derivative rules to calculate the derivative of SE2 563—1' y: We use the quotient rule with the Choices u = :32 and v = x3 — 1. Then we have
du d1} 2% = 2m and do; = 3x2 and so
91g _ d<§>
d3: .1 dx
vii—2:  ”23—:
1)2
(333  1mm)  (9658332)
_ ——(i:”——1)2—
_ 2x4 — 2:13 — 3x4
_ W
__ 7:134 ~ 23: 6. [15 points] Use appropriate derivative rules to calculate the derivative of 5
= 1 _ _
y ( x)
We use the generalized power rule with the choices to = 1 — i = 1 — 33—1 and n = 5.
d
Then we have _u_ z 0 — —x“2 = I” and so
dzc
dy _ (“a”)
dzv — dm
du
n—1_
— nu dac
1 4 72
= 1 _ _
5( 3:) x
5 1 4
= 7(1 — —) 7. Let y = 3:54 — 4:113. (a) [7 points] Find the critical values of x.
(b) [6 points] Make a sign chart for the derivative.
) (c [6 points] For each critical point, state whether it is a maximum, a minimum, or
neither. (a) We have 3/ = 3(4903) — 4(3132) : 1231:3 — 12232. To ﬁnd the critical :5 values we have
to solve the equation y’ = O for ac: 12$3—12$2=0
12$2(9:—1)=0
m2=00r(m—1):O
x=Oorzr=1. The critical :1: values are :c = 0 and :c = 1.
OD) cc \ Test Value 3/ = 12:103 — 12x2 Sign (—oo,0) —1 #24 —
0 (0,1) 0.5 —1.5 —
1 (1,00) 2 48 + (0) Since the derivative is negative before and after the critical point that has $ = 0,
this one is neither a maximum nor a minimum. Since the derivative is negative before
and positive after the critical point that has x = 1, this one is a (relative) minimum. 1
3. [12 points] Use the four—step process to ﬁnd the derivative of y = E 1
w + Am
1 1 1
ac + A55 —
_Z:_y $+A92 — %
A51: A30 y+Ay= Ay— ll H H H and so
d—y— lim Ay lim '1 — #1 ‘1
dg: _ Ago—>0 A3: Ana—>0 35(30 + Ax) Mm + 0) m2 ' Note: You know that the derivative of y = l = :E‘1 is going to turn out to be 1‘
y’ = —x#2 = —ﬁ because of stuff we have done since we discussed the fourstep process. This allows you to know whether or not you have made an error, but doesn’t
necessarily help to ﬁnd it if you have. 4. [12 points] The function y = Zﬁ — :1: has a critical point at (1,1). What conclusion
results from applying the second derivative test to this critical point? We start by ﬁnding the second derivative: 1
y/ : 2(ECE71/2) _ 1 : x—l/Q _ 1
1
I/ = __ —3/2
y 253 . At the critical point (1,1)7 Lt : l and 1 1
// _ __ —3/2 _ _ By the second derivative test, this critical point is a (relative) maximum. ...
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