<p>If f(x) = 2 for all real numbers x, then f(x+2) = ?
I need a clear explanation for why the answer is 2.</p>
<p>f(x) = 2 means that f(x) =y, therefore y=2. The graph of this function is a line parallel to x axis. This means that no matter what x value is, y stays the same.</p>
<p>Really cause I thought it was 4.
Edit: ^^Oh I get it now^^ That makes a lot of sense.</p>
<p>f(x) = 2 means that no matter what x value you select, the y value will be 2. (As AimingAt750 pointed out, this can graphed as the line y=2.) Thus, no matter what x is, f(x+2) will still be 2.</p>
<p>Think of a function as a machine. Depending on what you put into the machine you get a certain output. The function f defined by f(x) = 2 is the machine that always produces output 2 no matter what you put into it. So f(anything)=2. In particular, f(x+2)=2</p>
<p>Got it! Thanks. :D</p>
<p>f(x) + 2 is 4. f(x+2) remains at 2.</p>
<p>^Thank you!</p>
<p>Here’s another I couldn’t understand.</p>
<p>Which of the following statements is logically equivalent to: “If he studies, he will pass the course.”</p>
<p>A. He passed the course; therefore, he studied.
B. He did not study; therefore he will not pass the course.
C. He did not pass the course; therefore, he did not study.
D. He will pass the course only if he studies.
E. None of the above.</p>
<p>P.S. Are such questions common on SAT Math II?</p>
<p>Answer is C. Basically, the statement “A implies B” is logically equivalent to its contrapositive, “not B implies not A.”</p>
<p>You can’t claim that “not A implies not B” or “B implies A” unless you have a bijection (if and only if).</p>
<p>Such questions are pretty rare on Math II, but expect to see a question or so on it.</p>
<p>What wrong with the following calculation?</p>
<p>arcsin x = 2 arccos x
=> arcsin x cos x = 2 arccos x cosx
=> sin x arcsin x cos x = sin x . 2x
=> x cos x = 2x . sin x
=> sin x/cos x = 0.5
=> tan x = 0.5
=> x = arctan 0.5 and x = 0.464…</p>
<p>I don’t really understand what you’re doing here at all. It looks like you’re multiplying each side of the equation by sin x and then cos x. This does NOT simplify this expression in any way. The third equation is not equivalent to the second.</p>
<p>Here is a solution:</p>
<p>arcsin x = 2 arccos x
sin (arcsin x) = sin (2 arccos x) = 2sin(arccos x)cos(arccos x)
x =2sqrt(1-x^2)x
1/2 = sqrt(1-x^2)
1/4 = 1 - x^2
x^2 = 1 - 1/4 = 3/4
x = sqrt(3)/2 or x = -sqrt(3)/2</p>
<p>Notes: (1) In the second line I used the identity sin(2A) = 2sin A cos A
(2) To see that sin(arccos x)=sqrt(1-x^2), draw a right triangle, choose an angle theta, label the adjacent side x, the hypotenuse 1, and use the Pythagorean Theorem to get the opposite side to be sqrt(1-x^2).</p>
<p>arcsin x = 2 arccos x</p>
<p>This problem has a quick solution if you happen to remember one relationship:</p>
<p>Arcsin(x) + Arccos(x) = 90 degrees. If you want to know why this is true, remember that the cosine of an angle is equal to the sine of the complement. (The adjacent side of a given angle is the opposite side for the complementary angle.)</p>
<p>How does this help in this case?</p>
<p>Well, if arcsin x = 2 times arccos x we are looking for an angle that has a complement that is twice as big as the angle itself. Gotta be 30 degrees and 60 degrees. And x has to be cos(30) = root 3 over 2.</p>
<p>Where is this question from? If it is an SAT2 question, you would also have the answer choices to play with and you could solve it that way as well.</p>
<p>Wow this isn’t part of the SAT I math section right? I barely understood half of the stuff for the second question yellowcat429 posted! As for the first one, it’s fairly simple. If f(x)=2, then it doesn’t matter what value you pick for x, it will remain 2. So if you make x=3, then x+2 will equal 5. Since it says that f(x)=2 for all real numbers, it will stay 2 for x=5 as well. Just an example</p>
<p>@5am6996 the regular SAT math doesn’t include trig. You might see questions like those on Math II.</p>
<p>I understand your solutions. Thank you! 
One tiny doubt: What exactly makes my equations incorrect? If I multiply both sides of the equation by sin x, doesn’t the equality hold? My conclusion:
sin (arcsin x) = x and arcsin (sin x) = x, but
sin x arcsin x isn’t equal to x and neither is arcsin x sin x equal to x. True?</p>
<p>@yellowcat, here’s part of your solution:</p>
<p>=> arcsin x cos x = 2 arccos x cosx to
=> sin x arcsin x cos x = sin x . 2x</p>
<p>You essentially reduced “arccos x cos x” to “x.” This is incorrect.</p>
<p>sin(arcsin(x)) = x</p>
<p>arcsin(sin(x)) does not necessarily equal x.</p>
<p>Yellowcat, your conclusion is correct. sin (arcsin x) = x and arcsin (sin x) = x, but sin x arcsin x DOES NOT necessarily equal x. That was your error.</p>
<p>Ohh! Got it. :D</p>
<p>It’s easy to confuse composition of functions with product of functions. </p>
<p>Consider f(x) = x^2 and g(x) =sqr root(x)</p>
<p>f(g(x)) = (sqr root(x))^2 = x</p>
<p>But f(x) g(x) = (x^2)(sqr rt(x))=x^(5/2)</p>