<p>
[quote]
1. If c is a positive integer, the lim(x -> +inf) of (x^c)/(e^x) = ?
A) 0 B) 1 C) e D) c! E) DNE</p>
<ol>
<li>Suppose f and g are functions that are everywhere differentiable. Given that g is the inverse function of f and g(-2) = 5 and f '(5) = -1/2, then g '(-2) = ?</li>
</ol>
<p>A) 2 B) 1/2 C) 1/5 D) -1/5 E) -2</p>
<ol>
<li><p>If integral(1 to 4)[f(x)<em>dx] = 6, compute integral(1 to 4)[f(5-x)</em>dx]
A) 6 B) 3 C) 0 D) -1 E) -6
[/quote]
</p></li>
<li><p>A</p></li>
<li><p>Can you check this question for any mistakes? (Sorry, I just don't get it. I jump to the obvious answer: 0, which isn't a choice.</p></li>
<li><p>A</p></li>
</ol>
<p>sorry guys, i thought this thread died, should bump more often. anyways...</p>
<p>vader, that is correct, nice.</p>
<p>RESMonkey, yes A) -1 is the right answer. 3/2 is also correct, FTC FTW. I solved the x^2003 and x^(1/2003) one awhile ago but I remember using just one integral I think. Don't worry about this one the source is not from the AP test but it is noncalc.</p>
<p>You also got 1 and 3 correct. 2 is correctly stated. I'll give you a hint however. An inverse function f is one that returns the inputs of a different function g, so we can write f(g(x)) = x. It can be proved that if f is an inverse function of g, then g is also an inverse function of f. Hope this points you in the right direction.</p>
<p>ok, g = 1/f
g'(-2) = [f^-1(-2)]' = 1/f'(f^-1(x)), where f^-1(x) = -2; so since f^-1(x) = -2; and since its an inverse, the x and y are reversed, and g = f^-1, when y-variable = -2, its the x-variable of g, so y of f^-1 = x-of g, and you know g(-2) = 5, so f^-1(5) = -2. so it becomes: [f^-1(-2)]' = 1/f'(5), which is 1 / -1/2 = -2...so E</p>
<p>Only I converges, and you can use the alternating series test, right? Because II is always bigger than or equal to (1/n) which diverges since it's the harmonic series, and the integral of III is ln(ln(n)). But I'm probably wrong about this.</p>
<p>(a) a = f ' (x) > 0 @ t=2, so velocity is increasing. </p>
<p>(b) t = 12. when v is 55 again the Integration(0 to x) a(t) dt = 0; so the are under curve from 0 to 6 is 60; and are are from 6-10 is -30, and from 10-12 is -30; so 30+30-30-30 = 0; this is made possible at t=12..I'M SURE THERE IS ANOTHER MORE EFFICIENT WAY TO DO THIS THAN JUST GUESSAnd test: but it still still works!!! :)</p>
<p>(c)maximum occurs when f ' = 0; this happens at t=6, 16. BUT in order for it t0 be a max, the graph must switch from positive to 0 to negative; this happens at 6; but not at 16--so the answer is 6. Now to find the value you do Integration(0 to 6) a(t) dt; find the area under the curve its: 115 ft/s </p>
<p>(d) there is no such point becuase a (f ') is defined everywhere in the interval; if such a point existed, it would be at a point where a was undefined!</p>
<p>For part B, that is actually the most efficient way. If you actually look at the area of the a(t) curve from t=0 to 6, and then look at the area from t=6 to 12, you will realize that it's the exact same quadrilateral, and because t=0 to 6 is positive and t=6 to 12 is negative, they cancel each other out, which then gives you the conclusion that v(12) is also 55 ft/s.</p>
<p>@shravas</p>
<p>I do believe that 1 converges due to the alternating series test, but I also believe that 2 converges due to the ratio test.
let An = (1/n)<em>(3/2)^n
let A(n+1) = (1/(n+1))</em>(3/2)^(n+1)</p>
<p>lim |(1/(n+1))<em>(3/2)^(n+1)</em>1/((1/n)<em>(3/2)^n)| = lim|(2/3)</em>n/(n+1)|
using L'Hopital's rule, it then equals lim|2/3| as n approaches infinity, which is 2/3 < 1. Not sure about 3 though, I'm too tired to think :]</p>
<p>Nice job almost everything is right...except...in part (c) where you have to find the car's absolute maximum velocity...I think the area should be 60 right??? (Area from 0 to 6)</p>
<p>115 IS is the CORRECT ANSWER, I'll tell you why: You are correct, the area is 60, but you overlooked that initially at t=0, velocity was 55; so the total velocity is 55 + Integration(0 to 6) a(t) dt = 60 +55 = 115 :)</p>
<p>I think that 2 diverges, not converges, because the if you put it in the form of (a) * (r)^(n-1), then you get the ratio as 3/2, which is greater than 1, so it converges.</p>
<p>seperate the variables: dy / y = kt-----> integrate both sides ----->
ln[y] = kt + C ------>so y= Ce^(kt), where c is a constant, (b) is the only choice where this is the case :)</p>
<p>Yes that constant would. In essence you have to deal with two constants. </p>
<p>ln[y] = kt + C => y = e^(kt+C) = e^(kt) * e^C</p>
<p>The last step follows from properties of exponents. But e^C is another constant, so we can just call it K. Therefore y = Ke^(kt). You could also use C1 for the first constant and then C.</p>
<p>anyway new question:
what is the derivative of 2 apples? (seriously, no joke?, its an uber implicit diff question, a, p, l, e and s are all variables :))</p>