Math 161 Exam 2 Review
1. (a) x = 11
(b) x = 1 or x = 2
4. (a) ln x + 12 ln(1 + x2 )
(b)3 log2 x+ 21 log2 (x+1)−2 log2 (x−2)
2. (a) x = 125
(b) x = 27
(c) x = log104 3+7
(d) x = ln35
5. log4
3. (a)
3x
reflect about
y−axis
(b)
x−1
(x+1)4
(3x+1)3
log5 (2x−1)
2x
6. (a) x = 13 or − 13
(b) x = −6
(c) x = 1 or 3
shift down
by 2
/ 3−x
/ 3−x
7. (a) x =
(b) x =
−2
8. r =
1
4
3 ln 7+2 ln 4
ln 7−5 ln 4
ln 2−5 ln 3
4 ln 2−7 ln 3
8
5
ln
9. (a) k =
1
7
5
3
ln
(b) 300e 7 ln( 3 )
7 ln 3
(c) t = ln
( 53 )
10
(b)
ex
shift right
by 1
/ ex−1
reflect about
x−axis
/
shift up
by 3
−ex−1
/
−ex−1 + 3
10. t =
5
ln 2
0.087
11. (a) f 0 (x) = 12x2 − 12x + 1
1
−4
(b) f 0 (x) = −3x−2 + 3x− 2 + 15
2 x
9 x
8
0
(c) f (x) = 2 e + x
(d) f 0 (x) = x4 (x2 + 7x − 1) + (3 + 4 ln x)(2x + 7)
1
(c)
log2 (x)
shift left
by 2
/ log
2 (x
+ 2)
reflect about
x−axis
/ − log
2 (x
+ 2)
−1
2
3
√
12. y − 6 = 11(x − 2)
(d)
ln(x)
reflect about
y−axis
/
ln(−x)
shift up
by 3
/
− ln(−x) + 3
1
x
2
)−( x+e )(9x )
(e) f 0 (x) = ( 2 x )(1+3x
(1+3x3 )2
2
(f) f 0 (x) = 13 (8x4 − 8x2 + 3x)− 3 (32x3 − 16x + 3)
3
(g) f 0 (x) = 7ex +4x+9 (3x2 + 4)
3
1 √
9 − 21
(h) f 0 (x) = 9√x+8
+ 83 x− 2 )
3 x(2x
13. (a) abs. max:
abs. min:
(b) abs. max:
abs. min:
f (5) = 22;
f (3) = 2
f (−2) = 24;
f (−4) = −28
14. (a) x-int: 1,4; y-int: -16
(b) f 0 (x) = 3(x − 4)(x − 2)
(c) f increases on (−∞, 2) and (4, ∞). It decreases on (2, 4).
(d) It has a local maximum f (2) = 4 and a local min f (4) = 0.
(e) f 00 (x) = 6(x − 3)
(f) f is concave up on (3, ∞) and concave down on (−∞, 3).
It has a point of inflection at (3, 2).
(g)
y
(2,4)
(3,2)
(4,0)
0
1
2
3
x
4
-16
15. (a) x-int: 0,-2; y-int: 0
(b) f 0 (x) = 2x2 (2x + 3)
(c) f increases on (− 32 , ∞). It decreases on (−∞, − 32 ).
27
(d) It has a local min f (− 32 ) = − 16
.
00
(e) f (x) = 12x(x + 1)
(f) f is concave up on (−∞, −1) and (0, ∞).
It is concave down on (−1, 0).
It has point of inflections at (−1, −1) and (0, 0).
(g)
y
x
0
(-2,0)
(-1,-1)
(− 3
, − 27
)
2
16
2