Download Solving Quadratics, Dividing Polynomials Problem 1 Problem 2

Sept 20, 2011
MATH 140: Calculus I
Tutorial 2
Solving Quadratics, Dividing Polynomials
Problem 1
Solve for x: ln(x2 − 1) = 3.
Solution:
ln(x2 − 1) = 3
x2 − 1 = e3
p
x = e3 + 1
Problem 2
Solve for x: e2x − 3ex + 2 = 0.
Solution:
Let y = ex . Then we have a quadratic equation, y 2 − 3y + 2 = 0. By solving the quadratic equation, we
find that the roots are y = 2, 1. In other words, we must solve ex = 1 and ex = 2. But that means that
x = {0, ln(2)}.
Problem 3
Find the domain and range of the following function: f (x) =
Solution:
√
4 − x2 .
We know that 4 − x2 has to be positive in order to avoid imaginary numbers when taking the square root.
This happens for the domain x ∈ [−2, 2].
To determine the range, note that when x = 2, −2, f (x) = 0. Clearly f (x) reaches its maximal value when
x = 0; here f (x) = 2. So the range is [0, 2].
Problem 4
Find the domain and range of the following function: h(t) =
Solution:
4−t2
2−t .
Let’s first simplify:
h(t) =
4 − t2
(2 − t)(2 + t)
=
=2+t
2−t
2−t
There are no restrictions to the domain of this simplified function. The domain of that function is the whole
real line. However, consider the non-simplified function at t = 2. Here h(2) = 00 –this is undefined. So the
domain of h(t) is R \ 2 (the whole real line except for the point 2).
This missing point will affect the range. At t = 2, the simplified function has value 2 + 2 = 4. So the range
is R \ {4} (or, written differently, (−∞, 4) ∪ (4, ∞)).
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Sept 20, 2011
MATH 140: Calculus I
Tutorial 2
Problem 5
Complete the square: x2 + 3x.
Solution:
Let a be a real number. Then (x + a)2 = x2 + 2ax + a2 .
In our problem, 2a = 3, or a = 32 . So x2 + 3x = (x + 23 )2 − 94 .
Problem 6
Complete the square: x2 + 3x + 36.
Solution:
36 = 62 , so let’s first expand (x + 6)2 : x2 + 12x + 36.
In our case, x2 + 3x + 36 = (x + 6)2 − 9x.
Problem 7
Complete the square: 4x2 + 9x + 16.
Solution:
4x2 + 9x + 16 = 4(x2 + 94 x + 4). This is close to 4(x + 2)2 , so let’s expand this expression and see what we
need to take away.
4(x + 2)2 = 4(x2 + 4x + 4) = 4x2 + 16x + 16
For this problem, we find 4x2 + 9x + 16 = 4(x + 2)2 − 7x.
Problem 8
Complete the square: 5x2 + 2x + 15.
Solution:
5x2 + 2x + 15 = 5(x2 + 52 x + 3). But 5(x +
In this case, 5x2 + 2x + 15 = 5(x +
√
√
√
3)2 = 5x2 + 10 3x + 15.
√
3)2 + (2 − 10 3)x.
Problem 9
Solve the quadratic equation x2 − 4x − 16 = 0.
Solution:
Using the quadratic formula,
4±
√
16−4(−16)
2
√
= 2 ± 2 5.
Page 2 of 7
Sept 20, 2011
MATH 140: Calculus I
Tutorial 2
Problem 10
Solve the quadratic equation x2 + 3x − 4 = 0.
Solution:
Using the quadratic formula,
−3±
√
9−4(−4)
2
= −4, 1.
Problem 11
Find the roots of the following cubic x3 − 7x − 6 = 0, where one root is given to be −3.
Solution:
x2 + 3x + 2.
x−3
x3
− 7x − 6
3
2
− x + 3x
3x2 − 7x
− 3x2 + 9x
2x − 6
− 2x + 6
0
Problem 12
Find
x3 +3x−10
.
x+3
Solution:
x2 − 3x + 12.
x+3
x3
+ 3x − 10
− x3 − 3x2
− 3x2 + 3x
3x2 + 9x
12x − 10
− 12x − 36
− 46
So the solution is x2 − 3x + 12 −
46
x+3 .
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Sept 20, 2011
MATH 140: Calculus I
Tutorial 2
Solving Inequalities
Problem 13
Find where this inequality holds: x − 4 <
Solution:
p
(7 − x)(x − 2).
The right hand side of this equation is only valid where (7 − x)(x − 2) ≥ 0. So let’s find the range of x that
satisfies this inequality first. Clearly, the equation (7 − x)(x − 2) = 0 at x = 7, 2. So we can divide the real
line into three chunks: (−∞, 2), [2, 7], and (7, ∞). You can easily make a table or plug in 3 points from each
of these domains to see that (7 − x)(x − 2) ≥ 0 when x ∈ [2, 7].
Now we can search for the solution to the following inequality on [2, 7]:
(x − 4)2 < (7 − x)(x − 2)
p
However, we only care about when x ≥ 4 now. Why? Because x − 4 < 0 < (7 − x)(x − 2) when x < 4
(The left hand side is negative, and the right hand side is positive, positive numbers are always greater than
negative numbers, so the inequality is true!). We know that the range [2, 4) makes this inequality true. We
just want to find out where the inequality holds for x ∈ (4, 7].
Let’s find where the right hand side and left hand side are equal. To do this, we square both sides and add
the restriction that x ≥ 4. Luckily, we already have this restriction! So we compute:
(x − 4)2 = (7 − x)(x − 2)
x2 − 8x + 16 = −x2 + 9x − 14
2x2 − 17x + 30 = 0
Using the quadratic formula, the roots of this equation are at x = 2.5, 6. But we are only looking for
intersections in the interval [4, 7], so we only need to consider the point x = 6. By plugging in points (or
drawing the graph!), we see that the equality holds for x ∈ [4, 6), and does not hold above 6.
Combining this interval and the interval we found previously, we know the solution is x ∈ [2, 6).
Plot showing the inequality is true for [2,6)
x-4
((7-x)(x-2))1/2
4
3
f(x)
2
1
0
-1
-2
2.5
3
3.5
4
4.5
x
5
5.5
6
6.5
Page 4 of 7
Sept 20, 2011
MATH 140: Calculus I
Tutorial 2
Problem 14
This is Problem 1 from the sample quiz. Find for which x the following inequality is true: |3e2x − 5ex | < 2.
Solution:
We know that if |a| < b, then −b < a < b. Let’s expand our absolute value similarly:
−2 < 3e2x − 5ex < 2
Now let’s make a change of variables, y = ex , to make this a polynomial equation. With this change of
variables, we must be careful to specify that y > 0, as the exponential function (ex ) is never negative! Now
we have the following:
−2 < y(3y − 5) < 2
The equation y(3y − 5) has roots at y = 0, 53 . You can see this in the figures below as well. Between those
roots, the function is negative, and it’s positive elsewhere. In our case, there are two places that the function
and the line f (y) = 2 intersect, and two places that the function and the line f (y) = −2 intersect. We can
set y(3y − 5) equal to these values to find the points of intersection.
Graph of the inequality
5
Another way to visualize the inequality
4
2
3y -5y
2
-2
4
3.5
3
3
2.5
f(y)
2
f(y)
|3y2-5y|
2
1
2
1.5
0
1
-1
0.5
-2
-1
0
0
1
y
2
3
0
1
2
y
First let’s deal with y(3y − 5) = 2. Simplifying this, we find 3y 2 − 5y − 2 = 0, a quadratic equation. You
may factor this, or you can use the quadratic formula to solve for y:
p
5 ± 25 − 4(3)(−2)
y=
2·3
By simplifying this, one finds that y = 2, − 31 .
Similarly, set y(3y − 5) = −2. By simplifying we find 3y 2 − 5y + 2 = 0. Once again using the quadratic
equation, we find that y = 23 , 1. Therefore we conclude that this inequality is valid for y ∈ (− 31 , 23 ) ∪ (1, 2).
However we must add the original constraint that y must be positive. Therefore y ∈ (0, 23 ) ∪ (1, 2). Now
we can transform back to x, since if y = ex , then x = ln(y). The solution is that the inequality is valid for
x ∈ (−∞, ln( 23 )) ∪ (0, ln(2)). Convince yourself that ln( 23 ) = − ln( 23 ).
Page 5 of 7
Sept 20, 2011
MATH 140: Calculus I
Tutorial 2
Trig Functions
Problem 15
Convert from radians to degrees: 4π.
Solution:
◦
◦
4π( 180
π ) = 720 .
Problem 16
Find the other ratios (cos θ, tan θ): sin θ = 53 , 0 < θ < π/2.
Solution:
opposite
Since sin θ = hypotenuse
, we know that the side of the triangle opposite θ has length 3, and that the hypotenuse
has length 5. Call the missing side a (the side adjacent to θ). Then by the Pythagorean Theorem, a2 +32 = 52 ,
so a = 4. Therefore cos θ = 54 and tan θ = 34 .
Problem 17
Prove sin θ cot θ = cos θ.
Solution:
1
tan θ
cos θ
= sin θ
sin θ
= cos θ.
sin θ cot θ = sin θ
Problem 18
Prove sin x sin 2x + cos x cos 2x = cos x.
Solution:
Using an identity and that cos(x) is even:
1
1
[cos(x − 2x) − cos(3x)] + [cos(3x) + cos(x − 2x)]
2
2
= cos(−x)
sin x sin 2x + cos x cos 2x =
= cos(x).
Page 6 of 7
Sept 20, 2011
MATH 140: Calculus I
Tutorial 2
Problem 19
Solve for x: 2 sin2 x = 1, x ∈ [0, 2π].
Solution:
Taking the 2 to the right hand side and applying the square root we simplify to:
1
sin x = ± √
2
Solution is at the intersection
1.2
sin(x)
(1/2)1/2
1
-(1/2)1/2
0.8
0.6
f(x)
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
1
2
3
4
5
6
x
7π 11π
Clearly the solution is x = { π6 , π − π6 , π + π6 , 2π − π6 } = { π6 , 5π
6 , 6 , 6 }.
Final Note: To discuss Problem 12 from the homework, it’s best to come to office hours.
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