Unformatted text preview: Math 1110 Name: Homework 5 Due 9/29 or 9/30 in class Please print out these pages. Write your answers to the “text exercises” on separate paper and staple it to these pages. You should include
computational details. These problems will be assessed for completeness.
Always write neatly and legibly.
Please answer the “presentation problems” in the spaces provided.
Include full explanations and write your answers in complete, mathematically and grammatically correct sentences. Your answers will be assessed
for style and accuracy, and you will be given written feedback on these
problems. GRADES
Text exercises / 20 Pres probs / 20 Staple /1 Text exercises. Please do the following problems from the book.
§3.1 #8, 22, 23, 30, 34, 36
§3.2 #2, 8, 16, 27, 28, 29, 30, 36, 42, 46
Question 1. (10 points) Does knowing that a function g(t) is differentiable at t = 50 tell you
anything about the differentiability of the function −2g(t) at t = 50? Give reasons for your answer!
g(50 + h) − g(50)
Since g(t) is differentiable at t = 50 we know that lim
exists. We want to
h→ 50
h
−2g(50 + h) − (−2)g(50)
consider lim
. If this limit exists then −2g(t) is differentiable at t = 50.
h→ 50
h
Note:
−2g(50 + h) − (−2)g(50)
g(50 + h) − g(50)
= lim −2
h→ 50
h→ 50
h
h
−2g(50 + h) − (−2)g(50)
= −2 lim
h→ 50
h
lim −2g(50 + h) − (−2)g(50)
= g (50), and we know
h
that g is differentiable at t = 50. So we conclude that −2g(t) is differentiable at t = 50, and that its
derivative at that point is −2g (50).
We know that the ﬁnal limit exists because lim h→ 50 Math 1110 (Fall 2011) HW5 Presentation Problems 2 Question 2. (10 points) Find all points on the curve y = 2x3 + 3x2 − 12x + 1 where the tangent line
is horizontal. Give equations for all such tangent lines.
We want to ﬁnd points on the given curve where the derivative dy = 0, and then ﬁnd the tangent
dx
lines to the curve at these points. We begin by computing dy/dx.
dy
2(x + h)3 + 3(x + h)2 − 12(x + h) + 1 − (2x3 + 3x2 − 12x + 1)
= lim
dx h→ 0
h
3 + 3x2h + 3xh2 + h3) + 3(x2 + 2xh + h2) − 12(x + h) + 1 − (2x3 + 3x2 − 12x + 1)
2(x
= lim
h→ 0
h
2h + 3xh2 + h3) + 3(2xh + h2) − 12h
2(3x
= lim
h→ 0
h
= lim (2(3x2 + 3xh + h2) + 3(2x + h) − 12)
h→ 0
2 = 6x + 6x − 12 Now we plug in dy
dx = 0 and solve for x:
0 = 6x2 + 6x − 12
= 6(x2 + x − 2)
= 6(x − 1)(x + 2) So in order for dy to equal 0, we must have x = 1 or x = −2. Plugging x = 1 into the equation for
dx
our curve, we see that the only yvalue corresponding to x = 1 is 2(1)3 + 3(1)2 − 12(1) + 1 = −6.
Similarly, if x = −2, then y = 21. We want to ﬁnd the equation of a horizontal line going through
the point (1, −6) and an equation for a horizontal line that goes through the point (2, 5). The only
such equations are y = −6 and y = 21, respectively. Homework 5 Book Problem Answers
Section 3.1: Section 3.2: ...
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 '06
 MARTIN,C.
 Calculus, Derivative, Vector Space, lim

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