REVIEW QUESTIONS MATH 114 (SECTION C1): ELEMENTARY CALCULUS
Example 1. Find all x values satisfy 3x2 − 2x − 5 > 0. Given your answer in interval notation.
Example 2. Find the equations of the following lines.
(a) the line through (1, 2) and (4, 7);
(b) the line through (0, 1) that has slope 2;
(c) the line has x−intercept −2 and y−intercept 5.
Example 3. Let f (x) = √1x and g(x) = x2 − 9.
(a) Find f ◦ g(x) and g ◦ f (x)
(b) Find the domain and range of f (x), g(x), f ◦ g(x) and g ◦ f (x).
Example 4. Find the value of the following trigonometric functions
π
(a) sin 7π
12 (b) cos 12
Example 5. Find the following limits if they exist. If the limit does not exist, explain why.
2
−x−6
(a) limx−→3 xx2 −2x−3
√
2
x +7
(b) limx−→3 4−x−3
(c) limx−→0 sin2x4x
(d) limx−→∞ 2 sinxx+5
4x4 +3x3 −x2 +2x+5
(e) limx−→∞ 3x
4 +x3 +2x2 −6x−7
1
(f ) limx−→3 (x−3)
2
1
(g) limx−→4 4−x
Example 6. Find the vertical and horizontal asymptotes of f (x) =
Example 7. Find the oblique asymptote of f (x) =
3x2 −x+5
x2 −4 .
x2 +2x−6
.
x+1
Example 8. (a) State the definition of the continuity for function f (x) at the point x = x0 ;
(b) Findthe value of a and b that will make
 2x, x ≤ −2



f (x) =
ax2 − b, −2 < x < 4 continuous.




3x + 5, x ≥ 4
Example 9. Show that f (x) = x3 − 5x2 + 5 has at least one root in the interval [0, 2].
Example 10. (a) Use the definition of the derivative to find
not used.
(a) Find the equation of the line tangent to x2 at x = 3.
d 2
dx x .
No points will be given if the definition is
Example 11. Find the derivative of the following functions
(a) f (x) = sec(3x) − 5 tan(2x + 3)
(b) f (x) = x2 sin(2x2 − 3x)
2
(c) f (x) = √xx2 +3
−3
(d) f (x) = x +4x
x2
2
(e) f (x) = (3x − 6 sin(4x))8
3
2
Example 12. Find
d
dx y
for y defined through the equation 3x2 − 4xy + sin(xy) = 5.
Note: Examples 1-12 come from the review for midterm. I will not provide the solution
for those questions.
1
2
REVIEW QUESTIONS MATH 114 (SECTION C1): ELEMENTARY CALCULUS
Example 13. For f (x) = (1 + x)1/3 , find the linearization of f (x) (a) at x = 7, (b) at x = 0.
Recall: For y = f (x), the linearization of f (x) at a fixed point x0 its the tangent line to the
curve y = f (x) at x0 , that is L(x) = f (x0 ) + f ′ (x0 )(x − x0 ).
1
Solution (a) Let x0 = 7. f ′ (x) = 31 (1 + x)−2/3 , so f (7) = (1 + 7)1/3 = 2 and f ′ (7) = 13 (1 + 7)−2/3 = 13 14 = 12
.
1
′
Thus L(x) = f (x0 ) + f (x0 )(x − x0 ) = 2 + 12 (x − 7).
(b) We recall the known result: The linearization of f (x) = (1 + x)k at 0 is 1 + kx. Thus, the linearization
of (1 + x)1/3 at 0 is 1 + x3 .
Example 14. Find dy if y = x3 − 2x and the value of dy when x = 2 and dx = 0.1.
Recall: The differential of y = f (x) is defined by dy = f ′ (x)dx with dx is an independent variable.
So, for y = x3 − 2x, dy = (3x2 − 2)dx. When x = 2 and dx = 0.1, dy = (322 − 2)0.1 = 1.
Example 15. Find the absolute maximum and minimum values of f (x) on the given interval
(a) f (x) = x2 − 4x + 8 on [−1, 1], (b) f (x) = x1/3 − 16 x2 on [−1, 1].
Recall: If an interior x0 of the domain of f (x) satisfies f ′ (x0 ) or f ′ (x) is not defined at x0 ,
then x0 is called a critical point of f (x). To find the absolute maximum and minimum of f (x) on
[a, b], we only to find (1) all critical points of f (x); (2) the values of f (x) at these critical points
and endpoints a, b of [a, b]; (3) the largest value is the global maximum and the smallest value is
absolute minimum value of f (x).
Solution (a) f ′ (x) = 2x − 4, so f ′ (x) = 0 implies x = 2 which is not in [−1, 1]. Thus, f (x) has no critical
point. f (−1) = 1 + 4 + 8 = 13 and f (1) = 1 − 4 + 8 = 5. Therefore, the absolute maximum is 13 and minimum
is 5.
(b) can be solved similarly.
Example 16. (a) State Rolle’s Theorem and Mean Value Theorem,
(b) Use Mean Value Theorem to show that | cos a − cos b| ≤ |a − b| for all a and b.
Solution (a) Rolle’s Theorem: If f (x) satisfying (i) continuous on [a, b] (ii) differentiable on (a, b) (iii)
f (a) = f (b), then there is at least one c in (a, b) such that f ′ (c) = 0.
Mean Value Theorem: If f (x) satisfying (i) continuous on [a, b] (ii) differentiable on (a, b), then there is at least
(a)
one c in (a, b) such that f ′ (c) = f (b)−f
.
b−a
(b) Let f (x) = cos x then f (x) satisfies the conditions of Mean Value Theorem on [a, b](or[b, a]). Thus, there
is c between a and b such that
cos a − cos b
f ′ (c) =
.
a−b
Not that f (x) = cos x and f ′ (x) = (cos x)′ = − sin x. So, f ′ (c) = − sin c and |f ′ (c)| = | − sin c| ≤ 1. Therefore
| cos a−cos b|
≤ 1, that is | cos a − cos b| ≤ |a − b|.
|a−b|
Example 17. Use the First Derivative Test for Local extrema to identify the intervals on which f = x1/3 (x − 4)
is increasing and decreasing and find its local extreme values.
Recall: The First Derivative Test for Local Extrema: For a critical point c of f (x), (a) f ′ (x)
changes from negative to positive at c, then f has a local minimum at c; (b) f ′ (x) changes from
positive to negative at c, then f has a local maximum at c.
′
′
f ′ (x) = (x4/3 − 4x1/3 )′ = 43 x1/3 − 43 x−2/3 = 43 xx−1
2/3 . f (x) is not defined at 0 and f (x) = 0 implies x = 1.
Thus x = 0 and x = 1 are two critical points of f (x). These tow points divide (−∞, ∞) into three intervals
(−∞, 0), (0, 1), (1, ∞), the sign of f ′ (x) on these three intervals are −, −, +. So, on (−∞, 0) and (0, 1) f (s) is
decreasing and on (1, ∞) f is increasing. We can also see that at x = 1 the sign of f ′ (x) changes from − to +.
Thus f has a local minimum.
Example 18. Use the Second Derivative Test for Local extrema to find local extrema for f = x3 − 6x2 − 5.
REVIEW QUESTIONS MATH 114 (SECTION C1): ELEMENTARY CALCULUS
3
Recall: The Second Derivative Test for Local Extrema: Suppose that f ′′ is continuous on an
open interval that contains x = c. (a) If f ′ (x) = 0 and f ′′ (c) < 0, then f has a local maximum at
x = c, (b)If f ′ (x) = 0 and f ′′ (c) > 0, then f has a local minimum at x = c.
Solution f ′ (x) = 3x2 −12x = 3x(x−4). f ′ (x) = 0 implies x = 0 and x = 4. f ′′ (x) = 6x−12. f ′′ (0) = −12 < 0
and f ′′ (4) = 24 − 12 = 12 > 0. Thus, f has a local maximum f (0) = −5 at x = 0 and a local minimum
f (4) = −13 at x = 4.
Example 19. Use the Second Derivative Test for Concavity to identify where f = x4 − 4x3 is concave up and
concave down.
Recall: The graph of a differentiable function y = f (x) is (a) concave up on an interval I if f ′ (x)
is increasing on I (that is f ′′ (x) > 0) (a) concave down on an interval I if f ′ (x) is decreasing on I
(that is f ′′ (x) < 0)
Solution f ′′ (x) = 12x2 − 24x = 12x(x − 2). Then, f ′′ (x) = 0 implies x = 0 and x = 2. These two points
divide (−∞, ∞) into three intervals (−∞, 0), (0, 2), (2, ∞) on which the sign of f ′′ (x) are +, −, +. So, on (−∞, 0)
and (2, ∞), f is concave up, on (0, 2) f is concave down.
Example 20. A rectangle is to be inscribed in a semicircle of radius 4. What is the largest area the rectangle
can have.
√
Solution Let (x, 16 − x2 ) be the coordinates of the corner of the
√ rectangle obtained by placing the circle
and rectangle in the coordinate
plane.
The
height
of
the
rectangle
is
16 − x2 and the length is 2x. So
√
√the area
−2x2
′
2
√
of the rectangle is A(x) = 2x 16 − x . To maximize A(x) for 0 ≤ x ≤ 4, we find A (x) = 16−x62 + 2 16 − x2 .
√
√
√
2
2
2
A(x)√ = 0 implies
−2x
+
2(16
−
x
)
=
32
−
4x
=
0.
so,
x
=
2
2
or
x
=
−2
2.
Only
2
2 lies in [0, 4].
√ √
√
√ √
A(2 2) = 4 2 8 = 16, A(0) = A(4) = 0. Thus the area has maximum A(2 2) = 4 2 8 = 16.
Example 21. Find an antiderivative F (x) for f (x) = csc2 2x such that F (π/4) = 2.
)′
(
2
= csc2 (2x), F (x) = −1
Solution Since −1
2 cot(2x)
2 cot(2x) + C form general antiderivative of csc 2x.
−1
F (π/4) = 2 implies −1
2 cot(π/2) + C = C = 2. Thus F (x) = 2 cot(2x) + 2.
∫
∫
∫
∫
Example 22. Find (a)
sin 2xdx (b)
x1/4 dx (c)
sec2 (3x)dx (d)
cos2 xdx
∫
∫
∫
x1 + 14
−1
4 5
4 + C, (c)
sin 2xdx =
+
C
=
cos 2x + C, (b)
x1/4 dx =
x
sec2 (3x)dx =
1
2
5
1
+
4
∫
∫
∫
∫
1
1 + cos 2x
1
1
1
2
tan(3x) + C, (d) cos xdx =
dx =
1dx + cos 2xdx = x + sin(2x) + C.
3
2
2
2
2
Solution (a)
Example 23. Let f (x) = x2 .
(a) Find a formula for the Riemann sum obtained by dividing the interval [0, 4] into n subintervals and using
the right-hand endpoint for each ck ;
(b) Then, take a limit of the sum obtained in (a) to calculate the area under the curve over [0, 4].
4
4k
Solution The k−th subinterval is [ 4(k−1)
, 4(k)
n
n ]. So, ∆x = n ck = n and
)2
n
n (
n
∑
∑
∑
4(k)
64k 2
4
SP =
f (ck )∆x =
=
.
n
n
n3
k=1
k=1
k=1
(b)
A = lim
n−→∞
n
∑
k=1
f (ck )∆x = lim
n
∑
64k 2
n−→∞
Example 24. Evaluate the sums: (a)
k=1
n
∑
k=1
n3
n
64 ∑ 2
64
64 n(n + 1)(2n + 1)
=
.
k = lim 3
n−→∞ n3
n−→∞ n
6
3
= lim
k(2k + 3), (b)
k=1
40
∑
i=1
0.4
4
REVIEW QUESTIONS MATH 114 (SECTION C1): ELEMENTARY CALCULUS
Solution (a)
(b)
40
∑
n
∑
k(2k + 3) =
k=1
n
∑
2k 2 + 3k = 2
k=1
n
∑
k=1
k2 + 3
n
∑
k=
k=1
n(n + 1)(2n + 1)
n(n + 1)
+3
.
3
2
0.4 = 40 × 0.4 = 16.
i=1
Example 25. Show that 1 ≤
∫1
0
(1 + x4 )dx ≤ 2.
Consider f (x) = 1+x2 on [0, 1]. It has absolute maximum max f = f (1) = 2 and minimum min f = f (0) = 1.
∫1
∫1
So, min f (1 − 0) ≤ 0 (1 + x4 )dx ≤ max f × (1 − 0), that is 1 ≤ 0 (1 + x4 )dx ≤ 2.
∫ √2
∫ π
∫ π3
)
Example 26. Find (a)
+ 3 dx, (b)
(x + 2)dx, (c)
(1 + cos x)dx, (d)
2 sec2 (3x + 4)dx,
2
−2
1
0
0
∫ √2 2 √
∫ 3
∫ 1
x + x
3
1
(e)
dx,
(f
)
(t
−
2t)dt,
(g)
(x2 + 2)dx.
9
x2
1
−3
−1
∫
4
(x
Solution (a)
∫ 4(
x
−2
∫
∫ 4
)
1 4
1
42 − (−2)2
4 − (−2)
+ 3 dx =
xdx +
3dx = x2 |4−2 + 3x|4−2 =
+
= 6.
2
2 −2
4
4
2
−2
See another page for Solution of 26 (b)-(g) and 27-28.
(∫ 3
)
(∫ x 3
)
x
d
d
Example 27. Find (a) dx
(t + 2t)dt , (b) dx
sin(2t)dt .
a
a
Example 28. Find the area of the region enclosed by the curves y = 2 sin x and y = sin 2x over [0, π].