Math Functions Help!

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1

<p>Hi. So far I have averaging very low on the math portion of the sat, (First sat math section: 600, second sat math section: 560.) However, my reading and writing scores are much higher. After taking a few tests from the blue book, I have found out that I actually understand most of the concepts in the math section, but miss almost every single functions question! Obviously, I have found my issue. However, I have no idea how to solve it. This is where I hope you fellow CCers can help me out. :)</p>

<p>For example (I'll take a couple of problems from a blue book practice test I didn't do so hot on):</p>

<h1>4, section 2, practice test #5</h1>

<p>For which of the following functions is
f(-3)>f(3)</p>

<p>a. f(x) = 4x^2
b. f(x) = 4
c. f(x) = 4/x
d. f(x) = 4-x^3
e. f(x) = x^4+4</p>

<p>I don't even know how to begin this problem. I'm not exactly sure what it is even asking for. Do you plug in the -3 and 3 seperately to determine which one is greater? And if so, to where (the f(x) or the x) Any explanations?</p>

<p>Another functions question:</p>

<h1>13, section 4, test 5</h1>

<p>Let the function f be defined by f(x) = x+1.
If 2f(p)=20, what is the value of f(3p)?</p>

<p>Fill in the blank.</p>

<p>Again, no idea how to even start this. I don't know what to plug into what or how to set it up.</p>

<hr>

<p>I also have issues with problems like these, although I am not sure if they are considered functions problems:</p>

<h1>6, section 4, test 5</h1>

<p>If m and k are positive and 10m^2k^-1=100m,
what is m^-1 in terms of k?
a. k/10
b. k/90
c. square root of k/10
d. 1/10k
e. 1/90k</p>

<p>I really have no clue how to even start this question. Any help? How do you work this problem to where you get m^-1 alone?</p>

<p>Here's another example question I was confused about:</p>

<h1>14, section 2, practice test #5</h1>

<p>For how many ordered pairs of positive integers (x,y) is 2x+3y<6?</p>

<p>a. one
b. two
c. three
d. five</p>

<h2>e. seven</h2>

<p>Thanks to anyone who helps! I am really having issues and any suggestions or tips are appreciated!</p>

2

<h1>4 Intuitively, if you replace x = 3, f(x) becomes f(3). So replace x = 3, -3 into each of the functions and find which one satisfies f(-3) > f(3).</h1>

<p>A faster way to do it is note that choices A, B, and E contain even functions, that is, f(3) = f(-3), so these choices do not work. Answer is either C or D, quick plugging in yields D as the correct answer. //</p>

<h1>13 2f(p) = 20 → f(p) = 10. Since f(p) = p+1, p+1 = 10 → p = 9.</h1>

<p>Therefore f(3p) = f(27) = 27+1 = 28. //</p>

3

<h1>6, divide both sides by 100m^2:</h1>

<p>(1/10)k^(-1) = m^(-1)</p>

<p>(1/10)k^(-1) is equivalent to 1/(10k), D. //</p>

<h1>14 You can either brute-force, or graph the line 2x + 3y = 6 and count the number of lattice points contained within the first quadrant, bounded by 2x + 3y = 6. Turns out the only solution in positive integers (x,y) is (1,1) (since the other choices yield either x,y = 0 or 2x + 3y equals 6). The answer is A.</h1>

<p>Another way to do it is, because x and y are at least 1, 2x + 3y is at least 5. If you add any increment to x or y, you obtain a number at least 7, contradiction. Therefore the only solution is (1,1).//</p>