<p>Hi and thanks for opening this! :)
Here's the question:
Given f(2x+1) = x^2+3x+2, find f^-1(2) + f(3).
It seems it's easy, but my brain doesn't really want to think.
Please explain this to me, it'd be really apppreciated!
Thanks! :)</p>
<p>is the answer by any chance 3 or 6?
if yes, then i’ll explain cos i want to make sure what i did was right.</p>
<p>^Thank you xtanyax for paying attention!
Oh, I forgot to write them.
Unfortunately there’s no 3 or 6 
Here they are:
a) 5
b) 7
c) 8
d) 10
e) 11
Maybe try one more time with this answer choices, please? :)</p>
<p>i’m sorry. i can’t seem to get my head around this.</p>
<p>^No problem, thank you for trying 
Anybody else??</p>
<p>cud u however help me with this</p>
<p>f(x) =3 sin 3x and the domain of f(x) is b =< x =< b .If the graph of f crosses the x-axis exactly 7 times, which of the following could be b?</p>
<p>(A) pi/6
(B) pi/3
(C) 5pi/4
(D) 3pi/2
(E) 2pi</p>
<p>ans is (c)</p>
<p>thanks!</p>
<p>I’d graph it first.
So when we graph it we see that it crosses x at -4,-3,-2,-1,0,1,2,3,4.
And ans (c) is 5pi/4 which is approximately 3,9
maybe b should be btw 3 and 4 but I don’t know why, really, sorry.</p>
<p>that’s ok.</p>
<p>bumpppp anyone?</p>
<p>Yeah, I am looking at the b one and the explanation does not make any sense to me either… Weird…</p>
<p>I put the question in Yahoo answers, here it is.</p>
<p>[F(2x+1)</a> = x^2+3x+2, find f^-1(2) + f(3)? - Yahoo Answers](<a href=“Yahoo | Mail, Weather, Search, Politics, News, Finance, Sports & Videos”>Yahoo | Mail, Weather, Search, Politics, News, Finance, Sports & Videos)</p>
<p>Rogue (answer 2) answers this question.</p>
<p>Where did you get this question from? It looks like there are so many steps I can’t imagine this’d be on a 60-minute SAT exam!</p>
<p>Pope is right; f is not injective. Since f(1) = 2 and f(-5) = 2 (this can be seen by substituting x = 0 and x = -3 into the eq.), f^(-1) (2) is not defined. Rogue’s solution appears correct, but unnecessarily long.</p>
<p>In my opinion this is a terrible question.</p>
<p>I’m guessing you just factor the binomial and 2x+1 is a factor, which is the input.</p>
<p>Thank you everyone! I appreciate your help, but I still don’t get it :</p>
<p>Remember, f maps a set (in this case, the real #s) to another set (also the real #s). f^(-1) (2) basically asks, what number k is there such that f(k) = 2? In this case, there are two values of x (1, -5) such that f(k) = 2. Note that this is not the same as finding the values of x such that f(2x+1) = 2, but in this case there are also two values.</p>
<p>Since f^(-1) (2) can equal 1 or -5, it is not defined. That’s why I think it’s a terrible question.</p>
<p>Where did you find this?</p>
<p>I can’t wrap my head around it either.</p>
<p>Given f(2x+1) = x^2+3x+2, find f^-1(2) + f(3).
First we should find f(x).
Let us make a substitution t=2x+1. From here we get x=(t-1)/2. Substitute it into our function
F(2x+1)=f(t)=(t-1)^2/4+3(t-1)/2+2=(t^2-2t+1+6t-6+8)/4=(t^2+4t+3)/4
We can write x instead t: f(x)=(x^2+4x+3)/4. (1)
Next we should find the inverse function. From the definition we have y=f(x)⇔x=f-1(y). From the equation y=(x^2+4x+3)/4 we obtain:
X1=-2+\sqrt(1+4y) and x2=-2-\sqrt(1+4y). Therefore,
F^-1(y)=-2±\sqrt(1+4y). (2)
To find f^-1(2) + f(3) we take 2 instead y into (1) and 3 instead x into(2) and add them.
f^-1(2) + f(3)= -2±\sqrt(1+4<em>2)+(3^2+4</em>3+3)/4=-2+3+6=7 or =-2-3+6=1.
The answer: 7 or 1.</p>
<p>^I’m not sure if a substitution is the best method of finding f^(-1), since the algebraic steps you showed are unnecessarily long for an SAT question. Just use the fact that if x=0 or x=-3, you obtain f(2<em>0 + 1) = 2 and f(2</em>-3 + 1) = 2, so f^(-1) (2) = 1 or -5.</p>
<p>f(3) is easily found by plugging x=1, so f(3) = 6. So the answer is 7 or 1.</p>
<p>As I said, f is not injective (one-to-one) so no inverse function exists on R. Which is why I keep saying that this is not a good question.</p>