<p>For which of the following functions is it true that f(-x)= -f(x) for all values of x</p>
<p>(A) f(x)= x^2 +2
(B) f(x)= x^2 + 2x
(C) f(x)= x^3 + 2x
(D) f(x)= x^3 + 2
(E) f(x)= x + 2</p>
<p>Okay, so I already know how compute f(-x) ( Which is just making all the x's/whatever you plug in for "x" negative.) Idk how to compute the " -f(x) " part. </p>
<p>SPOILER : The answer is see</p>
<p>(A) f(-x)= -x^2 -2
(B) f(-x)= -x^2 - 2x
(C) f(-x)= -x^3 - 2x
(D) f(-x)= -x^3 - 2
(E) f(-x)= -x - 2</p>
<p>-f(x) is easily computed by just reversing all the signs of f(x).</p>
<p>(C) -f(x) = -x^3 - 2x</p>
<p>If you are not seeing it algebraically, try plugging in a specific example for x. So for each answer, you could evaluate say f(3) and f(-3). If the one of them is not the negative of the other, the answer is wrong.</p>
<p>This is NOT as fast as algebra. But it works. </p>
<p>What would be even faster would be to recognize that they were talking about “odd” functions and so you pick C as the only polynomial with only odd powers (linear and cubic).</p>
<p>when you have f(-x), you just put in -x everywhere there is an x, and simplify. When it is -f(x), it’s the same as doing say: -(5x+6), that is multiplying everything by -1. Just go through both sets, and see which one outputs to the same function.</p>
<p>That’s the slow way.</p>
<p>The fast way is to just check which function is symmetrical about the origin, and that’s your answer. And the only functions that follow this are ones with odd powers, cubic and linear, that eliminates A and B. Then you can eliminate E and D since they don’t go through with the origin, leaving you with C, which is your answer.</p>
<p>For example C is symmetrical on the origin: <a href=“f(x)=x^3+2x - Wolfram|Alpha”>f(x)=x^3+2x - Wolfram|Alpha, but the others are not: ex: <a href=“f(x)=x^2+2x - Wolfram|Alpha”>f(x)=x^2+2x - Wolfram|Alpha;
<p>Here is some theory in case you’re interested in that sort of thing:</p>
<p>A function f with the property that f(-x) = f(x) is called an even function. For example, f(x) = |x| is an even function because f(-x) = |-x| = |x| = f(x).</p>
<p>A function f with the property that f(-x) = -f(x) for all x is called an odd function. For example, g(x) = 1/x is odd because g(-x) = 1/(-x) = -1/x = -g(x).</p>
<p>A polynomial function is a function for which each term has the form ax^n where a is a real number and n is a positive integer. </p>
<p>Polynomial functions with only even powers of x are even functions. Keep in mind that a constant c is the same as cx^0, and so c is an even power of x. Here are some examples of polynomial functions that are even.</p>
<p>f(x) = x^2<br>
g(x) = 4<br>
h(x) = 3x^8 – 2x^6 + 9</p>
<p>Polynomial functions with only odd powers of x are odd functions. Keep in mind that x is the same as x^1, and so x is an odd power of x. Here are some examples of polynomial functions that are odd.</p>
<p>f(x) = x^3<br>
g(x) = x<br>
h(x) = 3x^11 – 2x^5 + 9x</p>
<p>A quick graphical analysis of even and odd functions: The graph of an even function is symmetrical with respect to the y-axis. This means that the y-axis acts like a “mirror,” and the graph “reflects” across this mirror. </p>
<p>The graph of an odd function is symmetrical with respect to the origin. This means that if you rotate the graph 180 degrees (or equivalently, turn it upside down) it will look the same as it did right side up. </p>
<p>So another way to determine if f(-x) = f(x) is to graph f in your graphing calculator, and see if the y-axis acts like a mirror. Another way to determine if f(-x) = -f(x) is to graph f in your graphing calculator, and see if it looks the same upside down. This technique will work for all functions (not just polynomials).</p>
<p>^Yes, for example, cos x is an even function because cos x = cos (-x).</p>
<p>In calculus and analysis, a “Taylor series” represents an infinitely differentiable function by an infinite series of polynomials (you won’t need to know what “differentiable” means for now). For cos x, it turns out that the Taylor series centered around 0 is</p>
<p>cos x = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + … (A Taylor series centered around 0 is sometimes called a Maclaurin series).</p>
<p>You can check that the series works by plotting cos x with the first few terms of the series, a very good approximation for points near x = 0. Since the RHS is an even function, the LHS will also be even.</p>
<p>@rspence, nice explanation, I like the taylor series mention, although you won’t need that on the test.</p>
<p>True, you don’t need Taylor series for SAT, but it’s kind of nice to know how these other functions such as sin, cos are “even” or “odd.”</p>
<p>Bringing dat calculus in.</p>