Review 1.1-2.8
Names:
1. Find the domain of the following:
(a) f (x) = 3x2 + 2x
Solution: R
(b) f (x) =
√
x+4
Solution: Need x + 4 ≥ 0, x ≥ −4. D = {x ∈ R : x ≥ −4} or D = [−4, ∞)
(c) f (x) =
p
√
5− x
Solution: We have a
i.e. x ≤ 25
D = [0, 25]
(d) f (x) =
√
√
√
x, which tells us x ≥ 0 We also need 5− x ≥ 0, 5 ≥ x, 25 ≥ x
√
x2 + 3
Solution:
Notice that x2 + 3 ≥ 3 for all x ∈ R. D = (−∞, ∞)
(e) f (x) =
x+2
x2 + 6x + 8
Solution:
We see x2 + 6x + 8 = (x + 2)(x + 4).
D = {x ∈ R : x 6= −4, −2} or D = (−∞, −4) ∪ (−4, −2) ∪ (−2, ∞)
(f) f (x) = √
3
4
x−5
Solution:
√
3
We know
that
x − 5 has D = (−∞, ∞). But since it’s in the denominator, we must
√
have 3 x − 5 6= 0, x − 5 6= 0, x 6= 5
D = {x ∈ R : x 6= 5}
2. State the base function and the transformations. Graph the following functions. Label any
important points. Draw dotted lines for any vertical or horizontal asymptotes.
(a) f (x) = x2 + 4x + 1
Solution: First we must complete the square: x2 + 4x + 4 − 3 = (x + 2)2 − 3 So x2 ,
shifted left by 2 and down by 3.
(b) g(x) = 3ex − 1
Solution: We stretch ex vertically by 3 and shift down by 1.
(c) f (x) = sin(x + π) − 2
Solution: We shift sin(x) to the left by π and down by 2.
(d) g(x) = | 12 x + 2|
Solution: Stretch |x| horizontally by 2 and shift left by 2.
1 1 √ √
(e) You should feel comfortable graphing xn , , 2 , x, 3 x, |x|, ex , ln(x), ax , all 6 trig funcx x
tions, piecewise functions.
3. State if the following are even, odd, or neither. Show your work to justify your answer.
(a) f (x) = x3 + sin(x)
Solution: f (−x) = (−x)3 + sin(−x) = −x3 − sin(x) = −(x3 + sin(x)) = −f (x). So
f (x) is odd.
(b) f (x) = x4 + 2x2 + 3
Solution: f (−x) = (−x)4 + 2(−x)2 + 3 = x4 + 2x2 + 3 = f (x). So f (x) is even.
(c) f (x) = 3x + cos(x)
Solution: f (−x) = 3(−x) + cos(−x) = −3x + cos(x). This isn’t f (x) or −f (x). So
f (x) is neither.
(d) f (x) = sin(cos(x))
Solution: f (−x) = sin(cos(−x)) = sin(cos(x)) = f (x). So this is even.
4. Evaluate the following
(a) sin π3
√
3
2
Solution:
(b) tan
3π
4
Solution: −1
(c) cos
5π
3
1
2
Solution:
(d) sin
π
2
Solution: 1
(e) tan
π
2
Solution: Undefined
√ (f) cos−1 23
π
6
Solution:
(g) arcsin
√ 2
2
π
4
Solution:
2
7
(h) tan arccos
Solution: Let√θ = arccos(2/7). So we have a triangle. So 22 +b2 = 72 , b =
3 5
So tan(θ) =
2
(i) sin sin−1 (x)
Solution: x
(j) cos sin−1
3
5
Solution: We have a 3 − 4 − 5 triangle. So cos(θ) =
4
5
√
√
45 = 3 5
5. Find the equation of a line for the following situations.
(a) The equation of a line that passes through the points (1, 3) and (−2, −3)
Solution:
−3−3
m = −2−1
= −6
−3 = 2
y − 3 = 2(x − 1) or y = 2x + 1
(b) The line perpendicular to the line y = 14 x + 7 that goes through the point (−1, 2)
Solution: m = −4
y − 2 = −4(x + 1) or y = −4x − 2
(c) The tangent line to the function f (x) = 2x2 − x at x = 2. Note: You are NOT allowed to
use derivative rules on the exam. (You can use them to check your work, but if you need
a derivative, you NEED TO use the limit definition).
Solution:
f (x + h) − f (x)
2(x + h)2 − (x + h) − (2x2 − x)
= lim
h→0
h→0
h
h
lim
2x2 + 4xh + 2h2 − x − h − 2x2 + x
h→0
h
= lim
= lim
h→0
h(4x + 2h − 1)
= lim 4x + 2h − 1 = 4x − 1
h→0
h
Now f 0 (2) = 4(2) − 1 = 7. And we have the point (2, f (2)) = (2, 6).
So y − 6 = 7(x − 2) or y = 7x − 8
6. Consider the functions f (x) =
√
x + 4 and g(x) = x2 − 3x.
(a) Find (f ◦ g)(x)
Solution:
f (g(x)) = f (x2 − 3x) =
p
x2 − 3x + 4
(b) Find (g ◦ f )(x)
Solution:
√
√
√
√
g(f (x)) = g( x + 4) = ( x + 4)2 − 3 x + 4 = x + 4 − 3 x + 4
(c) Find (f ◦ f )(x)
Solution:
√
f (f (x)) = f ( x + 4) =
q
√
x+4+4
7. Simplify the following expressions.
(a)
x5 x4
x3 x−2
Solution:
r
(b)
x9
= x8
x1
24 43
22
Solution:
r
24 (22 )3
=
22
r
24 26
=
22
r
210
= (28 )1/2 = 24
22
(c) log3 27
Solution:
3x = 27, x = 3
(d) ln(x) + 2 ln(x) − ln(x)
Solution:
ln(x) + ln(x2 ) − ln(x) = ln(x(x2 )) − ln(x) = ln(x3 ) − ln(x) = ln(x3 /x) = ln(x2 )
(e) log2
1
16
Solution: 2x =
1
16 ,
x = −4
8. Solve the following for x.
(a) log3 (x + 4) = 2
Solution: 3log3 (x+4) = 32
x+4=9
x=5
(b) ln(x − 8) = 0
Solution: eln(x−8) = e0
x−8=1
x=9
(c) ex
3 −3
=1
Solution: ln(ex
x3 − 3 = 0
x3 =√3
x= 33
3 −3
) = ln(1)
(d) 4x+7 = 16
Solution: log4 (4x+7 ) = log4 (16)
x+7=2
x = −5
9. Determine if there is an inverse of the function. If there is not, justify why there is not. If there
is, find the inverse.
(a) f (x) = x2 + 4.
Solution: f (x) is not 1-1 because f (1) = 5 = f (−1). So there is no inverse.
(b) f (x) = e7x+4
Solution: We know that f (x) is 1-1 because if f (x1 ) = f (x2 ), then e7x1 +4 = e7x2 +4
7x1 + 4 = 7x2 + 4
7x1 = 7x2
x1 = x2 .
The inverse:
y = e7x+4
ln(y) = 7x + 4
ln(y) − 4
x=
7
ln(x) − 4
−1
f (x) =
7
(c) f (x) = 4x + 8.
Solution: f (x) is 1-1 because: if f (x1 ) = f (x2 )
4x1 + 8 = 4x2 + 8
4x1 = 4x2
x1 = x2
The inverse:
y = 4x + 8
x = y−8
4
x−8
−1
f (x) =
4
10. Compute the following limits.
x−3
x→3 x2 − 9
(a) lim
Solution:
lim
x→3
(b)
lim
x→−1−
x−3
1
1
= lim
=
x→3
(x − 3)(x + 3)
x+3
6
x+3
x+1
Solution: We have #/0. Think −1.1. We see +/−
lim
x→−1−
x+3
= −∞
x+1
(h + 2)3 − 8
h→0
h
(c) lim
Solution:
h3 + 6h2 + 12h + 8 − 8
h(h2 + 6h + 12)
= lim
h→0
h→0
h
h
= lim
= lim h2 + 6h + 12 = 12
h→0
(d) lim ln(x)
x→∞
Solution: = ∞
(e)
2x2 − 4x
x→−∞ 3x3 + 6
lim
Solution:
2x2 /x3 − 4x/x3
2/x − 4/x2
=
lim
x→∞ 3 + 6/x3
x→−∞ 3x3 /x3 + 6/x3
= lim
=
0+0
0
= =0
3+0
3
(f)
lim
x→−2−
x2
2x + 4
+ 4x + 4
Solution:
= lim
x→−2−
2(x + 2)
2
= lim
=
(x + 2)(x + 2) x→−2− x + 2
# −2.1 +
=
= −∞
0
−
√
(g) lim
x→4
x−2
x−4
Solution:
√
√
x−2
x+2
x−4
√
= lim
= lim
·√
x→4 x − 4
x + 2 x→4 (x − 4)( x + 2)
= lim √
x→4
1
1
=
4
x+2
(h) lim x2 − 3x − 28
x→∞
Solution:
= lim (x − 7)(x + 4) = ∞
x→∞
1
2
−
(i) lim √
t→0 t 4 + t
t
Solution:
√
4 − (4 + t)
2− 4+t
√
= lim √
= lim √
t→0 t 4 + t(2 + 4 + t)
t→0 t 4 + t
−1
−t
√
√
= lim √
lim √
t 4 + t(2 + 4 + t) t→0 4 + t(2 + 4 + t)
t→0
=√
√
(j)
lim
x→−∞
−1
−1
−1
√ =
=
2(2
+
2)
8
4(2 + 4
4x2 + 7x
3x + 6
Solution:
√
p
4x2 /x2 + 7x/x2
4x2 + 7x/x
= lim
= − lim −
x→−∞
x→−∞ 3x/x + 6/x
3 + 6/x
p
√
4 + 7/x
4
2
= lim −
=−
=−
x→−∞
3 + 6/x
3
3
3
(k) lim x sin
x→0+
1
√
3
x
Solution:
1
≤1
−1 ≤ sin √
3
x
1
3
3
−x ≤ x sin √
≤ x3
3
x
Now: limx→0+ −x3 = 0 and limx→0− x3 = 0.
1
3
So by the Squeeze Theorem lim x sin √
=0
3
x→0
x
(l) lim exp
x→∞
5x2 + 2
4x3 − 2x
(Note: exp(x) = ex ).
Solution:
= exp
lim
x→∞
= exp
(m) lim arccos
x→−2
5x
2
−2x + 3x − 6
0
= exp(0) = 1
4
Solution:
= arccos
arccos
5/x + 2/x3
4 − 2/x2
lim
x→−2
5x
2
−2x + 3x − 6
(5(−2)
−10
π
= arccos
= arccos(1/2) =
2
−2(−2) + 3(−2) − 6
−8 − 6 − 6
3
x+3
− x − 12
(a) Find any vertical asymptotes. You must show why the possible points are vertical asymptotes.
11. Consider the function f (x) =
x2
Solution:
The possible points are when (x + 3)(x − 4) = 0, so x = −3, 4
lim
x→−3
1
1
x+3
= lim
=
(x + 3)(x − 4) x→−3 x − 4
−7
Not a VA
lim
x→4−
VA x = 4.
(b) Find any horizontal asymptotes.
x+3
= −∞
(x + 3)(x − 4)
Solution:
lim
x→∞ x2
x+3
1/x + 3/x2
= lim
=0
− x − 12 x→∞ 1 − 1/x − 12/x
x+3
1/x + 3/x2
=
lim
=0
x→−∞ x2 − x − 12
x→−∞ 1 − 1/x − 12/x
lim
HA: y = 0.
12. Redefine the function f (x) =
x−2
to be continuous x = 2.
x2 − 4
Solution:
lim
x→2
Let f (2) =
x−2
1
1
= lim
=
(x − 2)(x + 2) x→2 x + 2
4
1
4
13. Show that there is a jump discontinuity at x = −3 of the function
(
x+3
x ≤ −3
f (x) =
x+3
x > −3
x2 +5x+6
Solution:
lim x + 3 = 0
x→−3−
lim
x→−3+
1
1
x+3
= lim
=
= −1
(x + 3)(x + 2) x→−3+ x + 2
−1
Since the limit from the left and right are not equal, we have a jump discontinuity.
14. Show that
is continuous at x = 1
Solution:
• f (1) = 1 − 3 = −2
(
x−3
f (x) =
x2 − x − 2
x≤1
x>1
• lim x − 3 = 1 − 3 = −2
x→1−
lim x2 − x − 2 = 1 − 1 − 2 = −2
x→1+
So lim f (x) = −2
x→1
• lim f (x) = f (1)
x→1
15. Show that −|x + 3| − 4 is continuous on (−∞, ∞)
Solution: limx→a −|x + 3| − 4 = −|a + 3| − 4 = f (a).
16. Consider the function f (x) below.
(a) Evaluate the following.
i. lim f (x)
x→1+
Solution: 1
ii. lim f (x)
x→3−
Solution: 1
iii. lim f (x)
x→3+
Solution: 0
iv. lim f (x)
x→3
Solution: Does not exist
v. f (3)
Solution: 0
(b) State any points which are discontinuous. State as x, type.
Solution: x = −1, removable
x = 1, jump
x = 3, jump
x = 5, infinite
x = 7, removable
(c) State any points that are continuous but not differentiable.
Solution: x = −4, cusp
x = −2, corner
17. Prove there is a root of the function f (x) = x3 − 2x2 on [−1, 3]
Solution:
Since f (x) is a polynomial, f (x) is continuous.
f (−1) = (−1)3 − 2(−1)2 = −1 − 2 = −3
f (3) = (3)3 − 2(3)2 = 27 − 2(9) = 27 − 18 = 9
Since f (−1) = −1 < 0 < 9 = f (3) by IVT we have root.
18. Can we use IVT to show there is a root of f (x) =
answer.
x
x−2
on the interval [0, 4]. Justify your
Solution: No. The function is not continuous at x = 2. So we cannot use IVT.
19. Can we use IVT to show there is a root of f (x) = |x| on the interval [−2, 2]. Justify your
answer.
Solution: No. We have f (−2) = 2 = f (2). To use IVT we must have f (−2) 6= f (2)
20. Using the limit definition find f 0 (1) for the function f (x) =
√
x.
Solution:
√
√
√
f (h + 1) − f (1)
h+1− 1
h+1−1
f (1) = lim
= lim
= lim
h→0
h→0
h→0
h
h
h
√
√
h+1−1
h+1+1
h+1−1
= lim
·√
= lim √
h→0
h
h + 1 + 1 h→0 h( h + 1 + 1)
0
= lim √
h→0
1
1
=
2
h+1+1
21. Find the equation of the tangent line at x = −1 of the function f (x) =
Solution:
1
x3
1/(x + h)3 − 1/x3
f (x + h) − f (x)
= lim
h→0
h→0
h
h
f 0 (x) = lim
x3 − (x3 + 3x2 h + 3xh2 + h3 )
x3 − (x + h)3
=
lim
h→0
h→0 x3 (x + h)3 h
x3 (x + h)3 h
= lim
h(−3x2 − 3xh − h2 )
−3x2 − 3xh − h2
=
lim
h→0
h→0
x3 (x + h)3 h
x3 (x + h)3
= lim
=−
3
3x2
=− 4
6
x
x
Now f 0 (−1) = − 31 = −3 and f (−1) = 1, giving the point (−1, 1)
y − 1 = 3(x + 1).
22. Find the derivative of the function f (x) = 3x2 − x.
Solution:
3(x + h)2 − (x + h) − (3x2 − x)
3x2 + 6xh + 3h2 − x − h − 3x2 + x
= lim
h→0
h→0
h
h
f 0 (x) = lim
h(6x + 3h − 1)
= lim 6x + 3h − 1 = 6x − 1
h→0
h→0
h
= lim
So f 0 (x) = 6x − 1