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RE: [apcalculus] larson pg 137 exercise 69
Posted:
Oct 17, 2012 3:47 PM


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.  Dwayne,
Here is how I solved the problem that you asked.
step 1: we need to find the point that the tangent line and the graph intersect thus we need to set the function equal to the tangent line 5x4 = x^2 kx
Step 2: We need to find the derivative so we get f'(x) = 2x  k
Step 3: We know the value of the derivative at the intersection point is 5 so 5 = 2x  k or k = 2x  5
Step 4: Now lets plug in for k into step 1 so we get 5x  4= x^2 x(2x5) 5x 4 = x^2 2x^2 + 5x 4 = x^2 x = +/ 2
Step 5: Now lets plug in our two x values into step 3 to find our k values k = 2(2)  5 = 9 k = 2(2)  5 = 1
Take care
Douglas A. Dosky (c) 6142604699 email: ddosky@worthington.k12.oh.us Thomas Worthington High School AP Calculus BC and AB Teacher ________________________________________ From: Dwayne Wellington [mrdw27@gmail.com] Sent: Wednesday, October 17, 2012 9:29 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
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 To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus



