
#1
Jun2707, 02:41 PM

P: 4

here is my question, *thanks for helping* i need it badly
Consider the curve y = f(x) where f(x) = (x5)(2X+3) A) Show that P1 = (x1,y1) = (0,15) is on the curve and find and expression for the slope of the secant joining P1 with any other point P2 = (x2,f(x2)) , x2 cannot be 0 on the curve b) Show that there is always a point x*, lying between x1 = 0 and x2 such that the slope of the tangent line to the curve at x* is equal to the slope of the secant joining P1 and P2 ok and i am stuck on a) how would i go about solving it, would i take the derivate of the curve then sub the points from P1 to find P2? 



#2
Jun2707, 03:14 PM

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The secant line is just the line joining (x1,f(x1)) and (x2,f(x2)). There is no derivative involved  just a plain old slope.




#3
Jun2707, 03:14 PM

P: 927

For A) what is the value of f(x) when you plug in x=0? Edit: OK, a little too late posting




#4
Jun2707, 03:20 PM

P: 4

Help with secant question , thanks
okay so plan old slope i would find it by doing:
m = x2 x1 / f(x2)  f(x1) or slope of secant = f(0+h)  f(0) / h m = [(o+h)5][2(0+h)+3]  (5)(3) / h m = (h5)(2h+3) + 15 / h m = 2h^2 + 3h + 10h 15 +15 / h m= 2h7 ? 



#5
Jun2707, 03:32 PM

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The first, except it's upside down. It's deltay/deltax.




#6
Jun2707, 03:34 PM

P: 4

do i have enough info to complete the first, i dont really know how to go about doing it...




#7
Jun2707, 03:36 PM

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#8
Jun2707, 03:38 PM

P: 4

yeah i donty know x2




#9
Jun2707, 03:45 PM

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Just do it anyway. What's f(x2)? Use the definition of f.



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