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Difference Quotient Function


Date: 09/06/97 at 01:51:09
From: Nick Hogue
Subject: Difference quotient

Here's the question. Suppose you are asked to find the equation of the 
line tangent to a curve f(x) = x to the power of three minus seven x 
to the power of 2 plus twelve x plus 2 at the point (3,2) without 
using calculus. Suppose you have not taken calculus and work the 
problem without using derivatives. Describe how you would use the 
difference quotient function to find this equation, and then find the 
equation of the tangent line.


Date: 09/06/97 at 08:08:17
From: Doctor Jerry
Subject: Re: Difference quotient

Hi Nick,

Probably, f(x) = x^3-7x^2+12x+2. You want the equation of the tangent 
line at (3,2). I checked that f(3) = 2.

The slope of the line through (3,2) and the nearby point (3+h,f(3+h)) 
is

m(h) = (f(3+h) - f(3))/(3 - (3+h)) = (f(3+h)-f(3))/h.

If you calculate m(h) you'll find that m(h) = h^2+2h-3.  Now, try some 
small values of h, perhaps h = 0.1,0.01,0.001.  You'll find

   m(0.1) = -2.79
  m(0.01) = -2.9799
 m(0.001) = -2.998

It certainly appears that m(h) -> -3  as h->0. Now you can write the 
equation of the tangent line at (3,2).

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 09/06/97 at 19:56:19
From: Nick Hogue
Subject: Difference quotient

Okay, so what would the equation be?

f(x) = x^3-7x^2+12x+2 x>-3 or would I substitute 3 in for x 
or would I substitute x>-3 ?

Nick


Date: 09/12/97 at 20:18:48
From: Doctor Mike
Subject: Re: Difference quotient

Nick,   

When Dr. Jerry answered this he left the last step of finding the 
equation to you. What he was saying is that the slope of the tangent 
line to f at (3,2) is -3, so you need to find the equation of the one 
and only straight line with slope -3 which goes through the point 
(x,y) = (3,2).  

I know that if the slope is -3 then the equation of the line is 
y = -3*x+b where b is the y-intercept (the y coordinate where the line 
crosses the y-axis). So far we have only used the fact that the slope
is -3. To use the other fact that (x,y) = (3,2) is on this line
we substitute x = 3 and y = 2 into the equation of the line to get
   
       2 = -3*3 + b 
   
and solve for b = 11.  So the equation of the tangent line to
the function graph at (3,2) is  y = -3*x+11. I hope this helps.

-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Equations, Graphs, Translations

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