Mathematics Magazine for Grades 112 

2/2004 



There never was a good
war or a bad peace. Grade
12
Theory:
Vectors Definition: A vector
of dimension n is an ordered collection of n elements, which are called components. Notation: We often represent a vector by some letter, just
as we use a letter to denote a scalar (real number) in algebra. In typewritten
work, a vector is usually given a bold letter, such as A,
to distinguish it from a scalar quantity, such as A. In handwritten work, writing bold letters is difficult, so we typically just place a righthanded arrow over the letter to denote a vector. An ndimensional vector A has n elements denoted as A1, A2, ..., An. Symbolically, this can be written in multiple ways: A
= <A1, A2, ..., An> Example: (2,5), (1, 0, 2), (4.5), and (PI, a, b, 2/3) are all examples of vectors of dimension 2, 3, 1, and 4 respectively. The first vector has components 2 and 5. Solutions from the Previous Issue:Given the functions g(x) =
and h(x) =2x^{2} –3x + a , 1 Evaluate h(g(2)). Solution: g(2) = h(g(2))= 2(
)^{2 }3
+ a = 2∙2 3
+ a = 4 3
+ a 2 If , f(1) can only exist if g(1) = h(1). Find a so that f(1) exists. Solution: We know that g(1) = h(1) from the problem. Finding g(1) is easy: g(1) = Now, we know that h(1) = g(1), so h(1) = 2. In other words, if we plug in 2 for x in the expression called h, the result must be 2. This equation is very easy to set up, and finding a is as natural as chewing on tree bark (for those who find such things natural, that is). 2∙(1)^{2 } 3∙(1) + a = 2 2 – 3 + a = 2 a = 3 3 Evaluate the following delicious little limit: Solution: One of the fundamental properties of limits is that the limit of a sum is equal to the sum of the limits. In other words, you can evaluate this limit by calculating each individual term's limit as x approaches 0 and then adding up the results. You can evaluate the limit of the last two terms by direct substitution, but to do the first two, you have to remember these special limit cases: . You'll have to multiply the first term by 2/2 to get the denominator to be 2x (to match the numerator's x term and thus be able to apply the special limit property). In the second term, you can factor out a 1/3 to get the x terms to match:
You can further justify this answer by examining the graph of that funky mess. It clearly is heading toward a height of 3 when x = 0.
4
A function g(x) is defined as follows
Find the values of b and c which make g continuous. Solution: The graph of g is one seriously weird piece of garbage. It is linear until x equals 2 and then turns into a parabola until x = 4. After that, the graph is linear again.
We just have to design the parabola so that it connects those two segments together. To do that, we'll need to figure out what those two points are. Plug x = 2 into the first linear equation and x = 4 into the second. These are the correct lefthand limits and righthand limits respectively, which is why we don't have to worry about y = 2x  1 being defined at x = 4. y = 2∙(4) –1 y = y = 7 Therefore, we know the parabola must contain the points (2, ) and (4,7) or it will not meet up with the line segments correctly. We can plug each of these (x, y) coordinates into the parabola separately to get a system of equations: 2(2)^{2} + b(2) + c = 2∙(4)^{2} +b∙(4) + c = 7 8  2b + c = 32 + 4b + c = 7 2b + c = 4b + c = 25 You can solve this system just like any system you did in beginning Algebra (although the answer will be a little stranger than beginning algebra problems, I grant you). The quickest way is to graph both lines on the calculator and calculate the intersection point. The final answer is b = , c = , and you can verify this graphically by plugging the values into the parabola. The pieces should match up exactly, creating a continuous function.
Proposed Exercises:Proposed by Mihai Rosu Professor 1 Prove that the function is a constant function for every . 2 Prove the identity. 3 Prove the identity . 4. Prove that . 
