Thursday, December 3, 2009

Another problem... it counts as a quiz... if you can't get it within 3-4 hours then nevermind..

another problem... it counts as a quiz... if you can't get it within 3-4 hours then nevermind...



quadratics having a common tangent



x^2 + ax + b and y = cx -x^2 have a common tangent line at the point (1,0). Find a, b, and c.



--------------------



problem 2



find dy / dt



y= 4 sin ( sqrt (1+ sqrt (x) )



--------------------



find value of the derivative of (f o g) @ the given value of x. f(u) = 2u/u^2 + 1 , u = g(x) = 1/x^2 - 1 , x = -1



-------------------



Another problem... it counts as a quiz... if you can't get it within 3-4 hours then nevermind...?

f(x) = x^2 + ax + b , g(x) = cx - x^2 have a common tangent line at the point (1,0). This means



f'(1) = g'(1)



Let's solve for f'(x) and g'(x).



f'(x) = 2x + a. g'(x) = c - 2x



f'(1) = 2(1) + a = 2 + a



g'(1) = c - 2(1) = c - 2



Since f'(1) = g'(1), then



2 + a = c - 2



It also follows that they both go through the point (1,0), so



f(1) = 0 = 1^2 + a(1) + b = 1 + a + b



1 + a + b = 0



g(1) = 0 = c(1) - 1^2 = c - 1 = 0



c - 1 = 0



Three equations, three unknowns:



2 + a = c - 2



1 + a + b = 0



c - 1 = 0



The third equation implies c = 1. This means we can get "a" from the first equation.



2 + a = 1 - 2



2 + a = -1



a = -3



And we can get b from the second equation, given a = -3 and



c = 1.



1 + (-3) + b = 0



-2 + b = 0



b = 2



a = -3, b = 2, c = 1



Another problem... it counts as a quiz... if you can't get it within 3-4 hours then nevermind...?

This will take me about ten min. But why should I do your work and you get quiz credit. That is pretty lazy isn't it?



Another problem... it counts as a quiz... if you can't get it within 3-4 hours then nevermind...?

do your own work! how else are you gonna learn?



Another problem... it counts as a quiz... if you can't get it within 3-4 hours then nevermind...?

Solving 1 for U



The point (1,0) verifies every eq.



So 1+a+b=0 --%26gt; a+b= -1



%26amp; c-1 = 0 --%26gt; c=1



The slope of the tangent is the samt at the point (1,0) for the two curves



2x + a = c -2x at(1,0)



i.e 2+a = c - 2 = -1



So a = -3 , b =2 , c=1

No comments:

Post a Comment

 
mortgage loan