Difference Quotient FunctionDate: 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/ |
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