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How do i decide theese curves?

 Jan 11, 2019
 #3
avatar+18467 
+1

The value of f' at x = 2 = slope of the curve at that point....

slope of tangent line 1 =  4/2     slope of line 2 = - 1/3

slope 1 * slope 2 = -4/6 = -2/3  

 Jan 11, 2019
 #4
avatar+5232 
+1

Now see this is a math question.  No need for the off-topic flag here.

 Jan 11, 2019
 #5
avatar+102465 
+1

Maybe guest put it there by mistake. Anyway, it is gone now :)

Melody  Jan 12, 2019
 #6
avatar+7685 
+1

In both diagrams, the tangent line is given. By definition of derivative, the derivatives of the functions at the point of tangency is just the slope of tangent line.

 

Therefore, by calculating(or approximating? :D) the slope of the tangents, we get \(p'(2) = 2\) and \(q'(2) = \dfrac{-1}{3}\).

 

And then, we see the graph of p(x) passes through (2,3), and the graph of q(x) passes through (2, -1). So p(2) = 3 and q(2) = -1.

 

By product rule of differentiation:

\(\quad r'(2)\\ = p(2) q'(2) + q(2) p'(2)\\ = 3\cdot \dfrac{-1}{3} + (-1)\cdot 2\\ = -1 - 2\\ = -3\)

There we go :)

 Jan 14, 2019

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