<p>I had this question in an exam and do not know how to find the answers. Does anyone know how to do this?</p>
<p>find (f-1)'(3) for f(x) = sqrt (x^3 + x^2 + x + 6)</p>
<p>By (f-1)' , I mean derivative of the inverse function...</p>
<p>Thanks...</p>
<p>One of the parts of this process is to find the rule for f-inverse-prime as a function of f. Some people memorize this, others don't as the rule is somewhat easily enough derived on the spot using implicit differentiation.</p>
<p>If y = f (x), then x = f-inverse of y. Then taking the derivative of both sides yields 1 = f-inverse-prime of y * y'. So f-inverse-prime of y = 1/(y'). Keep in mind here that y' = f '(x), so that when we evaluate y' later, we're evaluating it for a given value of x, while the 3 value we're given is a value of y.</p>
<p>This means we need to find a corresponding x-value for when y = 3. So 3 = sqrt (x^3 + x^2 + x + 6), and squaring both sides yields 9 = x^3 + x^2 + x + 6 which suggests x^3 + x^2 + x = 3 and examination suggests x = 1 as a possible result. (If this is a calculator test, you might search for other solutions to this equation using graphs, although the existence of other solutions would imply that f-inverse is not a function and would create multiple solutions. In the absence of a calculator, you really don't have any great techniques besides synthetic/polynomial division to solve x^3 + x^2 + x - 3 = 0 by hand, and then to factor the remaining quadratic. I do not have a calculator handy while asking this, nor the desire to determine whether other solutions exist, so I'll leave this task for you.)</p>
<p>Now we find y' for your function, using the chain rule to arrive at (1/2)(x^3 + x^2 + x + 6)^(-1/2)(3x^2 + 2x + 1).</p>
<p>So y'(1) = (1/2)(9)^(-1/2)(6) = (1/2)(1/3)(6) = 1.</p>
<p>And (f-inverse)'(3) = 1/(y'(1)) = 1.</p>
<hr>
<p>As an interesting little side note here... many teachers who frequently find that they have well-prepared students for the AP Calculus exam got burned last year on a free-response question, where they asked students to calculate f-inverse-prime. For AP Calc students everywhere, I'd expect a bit of an overkill, over-response from teachers nationwide who are up on these kinds of things.</p>
<p>thank you for the thorough explanation...</p>
<p>mathprof, you were of tremendous help and the solution is obvious to me.</p>
<p>I have the last two questions on that exam that I had/still have trouble with...</p>
<ol>
<li><p>solve the equation for x : ln(1+ e^(-x)) = 3.
-is exponentiating both sides the right thing to do?</p></li>
<li><p>lim as x approaches positive infinity of (3e^(2x)-1)/(2e^(2x)+1)</p></li>
<li><p>won't the two quantities always differ by the same amount? I don't understand why the answer is 1.</p></li>
</ol>
<p>I appreciate if anyone can help me with those...</p>
<ol>
<li><p>It is, followed by solving for x by subtracting 1 from both sides, taking the natural logarithm of both sides and dividing both sides by -1. It doesn't appear that you get a nice clean answer for this question.</p></li>
<li><p>Is there a typo in this question? I'm actually thinking the answer to the question, as written, is 3/2.</p></li>
</ol>
<p>Before we tackle your question, let's take a look at a simpler question, of the lim as x approaches positive infinity of (3x^4 + 3)/(2x^4 - 3). How do we know the answer to these kinds of questions? Some people remember it based on the idea of the degree of the numerator versus the denominator.</p>
<p>But another way to look at this is by dividing each term by the largest "degree" term that we see. So in this case, we divide each of the terms by x^4. So an equivalent question is finding the limit as x goes to infinity of ((3x^4)/(x^4) + 3/(x^4))/((2x^4)/(x^4) - 3/(x^4)) = (3 + 3/x^4)/(2 - 3/x^4). Now as x goes to infinity, 3/x^4 goes to 0, so the numerator heads towards 3 and the denominator heads towards 2. It's completely irrelevant that the numerator is always slightly larger than 3, and that the denominator is always slightly less than 2. In the long run, those little bits slightly larger than 3 are less and less significant, as is those little bits slightly less than 2.</p>
<p>Now, let's go back to the original question. In this case, the divisior of choice is e^(2x), and the result should be 3/2, and for the same reason.</p>
<p>I understand that concept, but didn't realize it could be applied here. Thanks once again for the great help.</p>
<p>
[quote]
As an interesting little side note here... many teachers who frequently find that they have well-prepared students for the AP Calculus exam got burned last year on a free-response question, where they asked students to calculate f-inverse-prime. For AP Calc students everywhere, I'd expect a bit of an overkill, over-response from teachers nationwide who are up on these kinds of things.
[/quote]
</p>
<p>Ah I remember that. My Calc teacher went over that for like 2 sec... she said that it hadn't come up on an AP exam for a long time, but she would teach it to us anyway just so that we'd know. I still got that problem wrong though.</p>
<p>Sorry for getting OT...</p>
<p>
[quote]
Ah I remember that. My Calc teacher went over that for like 2 sec... she said that it hadn't come up on an AP exam for a long time, but she would teach it to us anyway just so that we'd know. I still got that problem wrong though.
[/quote]
</p>
<p>The same thing happened to me with my students.</p>
<p>IMO, the f-inverse-prime type equations aren't worth memorizing, because you can rederive them easily enough and because they are rare enough. As soon as we all overreact to this one, we'll not see it again until 2028. :)</p>
<p>the average score for that question on the AP was .96 out of 9.
60% of students didnt get any points. just shows how unprepared everyone was for that question. i think i got a couple points on it myself though :)</p>