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.
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problem 2
find dy / dt
y= 4 sin ( sqrt (1+ sqrt (x) )
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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
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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
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