Multivariable Calculus 9th Edition by James Stewart solution manual

6. The denominator cannot equal 0, so 1 − tan 6= 0 ⇔ tan 6= 1 ⇔  6=

4
+ . The tangent function is not defined
if  6=

2
+ . Thus, the domain of () =
1
1 − tan
is
  |  6=

4
+ ,  6=

2
+ ,  an integer  .
7. (a) An equation for the family of linear functions with slope 2
is  = () = 2 + , where  is the ­intercept.
(b) (2) = 1 means that the point (21) is on the graph of . We can use the
point­slope form of a line to obtain an equation for the family of linear
functions through the point (21).  − 1 = ( − 2), which is equivalent
to  =  + (1 − 2) in slope­intercept form.
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 FUNCTIONSAND MODELS
(c) To belong to both families, an equation must have slope  = 2, so the equation in part (b),  =  + (1 − 2),
becomes  = 2 − 3. It is the only function that belongs to both families.
8. All members of the family of linear functions () = 1 + ( + 3) have
graphs that are lines passing through the point (−31).
9. All members of the family of linear functions () =  −  have graphs
that are lines with slope −1. The ­intercept is .
10. We graph  () =  3 −  2 for  = −2, 0, 1, and 3. For  6= 0,
() =  3 − 2 =  2 (−) has two ­intercepts, 0 and . The curve has
one decreasing portion that begins or ends at the origin and increases in
length as || increases; the decreasing portion is in quadrant II for   0 and
in quadrant IV for   0.
11. Because  is a quadratic function, we know it is of the form () =  2 +  + . The ­intercept is 18, so (0) = 18 ⇒
 = 18 and () =  2 +  + 18. Since the points (30) and (42) lie on the graph of , we have
 (3) = 0 ⇒ 9 + 3 + 18 = 0 ⇒ 3 +  = −6 (1)
 (4) = 2 ⇒ 16 + 4 + 18 = 2 ⇒ 4 +  = −4 (2)
This is a system of two equations in the unknowns  and , and subtracting (1) from (2) gives  = 2. From (1),
3(2) +  = −6 ⇒  = −12, so a formula for  is () = 2 2 − 12 + 18.
12.  is a quadratic function so () =  2 +  + . The y­intercept is 1, so (0) = 1 ⇒  = 1 and  () =  2 +  + 1.
Since the points (−22) and (1−25) lie on the graph of , we have
(−2) = 2 ⇒ 4 − 2 + 1 = 2 ⇒ 4 − 2 = 1 (1)
(1) = −25 ⇒  +  + 1 = −25 ⇒  +  = −35 (2)
Then (1) +2 · (2) gives us 6 = −6 ⇒  = −1 and from (2), we have −1 +  = −35 ⇒  = −25, so a formula for
 is () = − 2 − 25 + 1.
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2 MATHEMATICAL MODELS:A CATALOG OF ESSENTIALFUNCTIONS
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13. Since (−1) = (0) = (2) = 0,  has zeros of −1, 0, and 2, so an equation for  is () = [ − (−1)]( − 0)( − 2),
or () = ( + 1)( − 2). Because (1) = 6, we’ll substitute 1 for and 6 for ().
6 = (1)(2)(−1) ⇒ −2 = 6 ⇒  = −3, so an equation for  is () = −3( + 1)( − 2).
14. (a) For  = 002 + 850, the slope is 002, which means that the average surface temperature of the world is increasing at
a rate of 002
◦ C per year. The ­intercept is 850, which represents the average surface temperature in ◦ C in the
year 1900.
(b)  = 2100 − 1900 = 200 ⇒  = 002(200)+ 850 = 1250
◦ C
15. (a)  = 200, so  = 00417( + 1) = 00417(200)( + 1) = 834 + 834. The slope is 834, which represents the
change in mg of the dosage for a child for each change of 1 year in age.
(b) For a newborn,  = 0, so  = 834 mg.
16. (a) (b) The slope of −4 means that for each increase of 1 dollar for a
rental space, the number of spaces rented decreases by 4. The
­intercept of 200 is the number of spaces that would be occupied