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Topic:
[apcalculus] larson pg 137 exercise 69
Replies:
1
Last Post:
Oct 17, 2012 11:18 AM




RE: [apcalculus] larson pg 137 exercise 69
Posted:
Oct 17, 2012 11:18 AM


NOTE: This apcalculus EDG will be closing in the next few weeks. Please sign up for the new AP Calculus Teacher Community Forum at https://apcommunity.collegeboard.org/gettingstarted and post messages there.  Hello,
Since the line is tangent to the graph, you know two things: 1. the graphs must be equal at some xvalue and 2. their derivatives must also be equal at that same xvalue. So, x^2kx=5x4 and 2xk=5. Thus, k=2x5 from the second equation. Sub that into the first equation for k to get x^2(2x5)x=5x4. Simplify to get x^2=4, so x=+/2. Since you now have xvalues where it is possible, you can find the corresponding k values by using the k=2x5 equation. HTH.
Steve Becker Math Teacher (503) 7857948 "What lies behind us and what lies before us are tiny matters compared to what lies within us."  Oliver Wendell Holmes
________________________________________ From: Dwayne Wellington [mrdw27@gmail.com] Sent: Wednesday, October 17, 2012 6:29 AM To: AP Calculus Subject: [apcalculus] larson pg 137 exercise 69
NOTE: This apcalculus EDG will be closing in the next few weeks. Please sign up for the new AP Calculus Teacher Community Forum at https://apcommunity.collegeboard.org/gettingstarted and post messages there.  Good morning, I was working on exercise 69 from Larson early Transcendentals and am missing the key to answering the question.
Find the value of k such that the line is tangent to the graph of the function.
(function) f(x) = x^2  kx (line) y = 5x 4
the answers are x= 1, 9 but I can't seem to get that when I work out the problem. Thank you for the assistance in advance.  To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus
 To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus



