MA 136 Fall 2020
Review Problems for Test # 3
dy
of the functions in questions 1 – 13.
dx
y  ln(3x)
Find
1.
2. y  ln( x3  4 x  6)
 x 32 1  x4
3. y  ln 
 x3  2

4. y  ln(cos e x )
5. y  e2 x sin 4 x
6. y  e3 x ln 5 x




7. y  x3 ln x 2
8. y  x3 (ln x)2
9. y  2 xe5 x
2
x
10. y  e
11. y  (sin 1 ( x 2 ))3
1
12. y  tan 1 (2 x)
3
13. y  5 tan 1 (3e2 x )
In questions 14 and 15 use implicit differentiation to find
14. sin( x  y )  1  2 x
15. 3x 2  2 xy  y 2  4 y  2
dy
dx
Find
(a) x- coordinates of all critical points
(b) interval(s) on which f is increasing
(c) interval(s) on which f is decreasing
(d) intervals on which f is concave up
(e) intervals on which f is concave down
(f) x- coordinates of the inflection points of f
and (e) locate all relative maxima and relative minima of the following functions (in questions
16 – 18):
16. f ( x )  27 x  x 3
17. f ( x)  x 3  x 3
x
18. f ( x)  2
x 2
4
1
Locate the critical points of functions in questions 19 – 21.
x2
19. f ( x)  3
x 8
2
20. f ( x)  x 2 ( x  1) 3
1
21. f ( x)  x 
x
Find the absolute extrema of the following function on the given closed intervals
3
22. f ( x)  x 3  x 2 on [ - 1, 2].
2
23. f ( x)  3cos( x) on [0, 2 ].
24. f ( x)  5e x  e2 x on [ - 1, 2].
Answers.
1
1.
x
3x 2  4
2. 3
x  4x  6
3
4 x3
3x 2
3.


2 x 2(1  x 4 ) x3  2
4. e x tan e x
5. 2e2 x (sin 4 x  2cos 4 x)
6. e3 x (3ln 5x  1/ x)
7. x 2 (3ln x 2  2)
8. x 2 ln x(3ln x  2)
9. 2e5 x (1  5 x)
2 2x
e
x2
6 x(sin 1 ( x 2 )) 2
10. 
11.
12.
13.
14.
15.
17.
1  x4
2
3(1  4 x 2 )
30e2 x
1  9e4 x
dy 2  cos( x  y )

dx
cos( x  y )
dy
2 y  6x

dx 2 x  2 y  4
4
1
4
2
Hint: f ( x)  x1/ 3  x 2 / 3 and f ( x)  x 2 / 3  x 5/ 3
3
3
9
9
18. Hint: f ( x) 
 x2  2
and
( x 2  2)2
f ( x) 
2 x3  12 x
( x 2  2)3
19. -2 and 2 3 2
20. 0, 1 and ¾
21. -1, 1 and 0
22. Abs. min. of -5/2 at x = - 1 and Abs. max. of 2 at x = 2.
23. Abs. min. of - 3 at x = π and Abs. max. of 3 at x = 0 and x = 2π.
24. Abs. min. of 5e 2  e 4 at x = 2
and Abs. max. of 25/4 at x = ln(5/2).