Calculus Easy Slope of Curve Problem

by tensirk
Tags: calculus, curve, secant, slope
 P: 12 1. The problem statement, all variables and given/known data Find the slope of the curve for the given value of x. y=x3-8x; x=1 a. the slope is -3. b. the slope is 1. c. the slope is -5. d. the slope is 3. 2. Relevant equations Would it be... Vav= s(t)-s(a)/t-a? 3. The attempt at a solution I know this is a really simple problem, but I just can't figure it out and my book is no help at all. I know this problem deals with the slope of a secant line and I think I am supposed to make a table of values to plug into the equation of a secant line. Help is MUCH MUCH MUCH appreciated. This is not a problem that will be graded for homework. I have a Calculus test soon and have seemingly forgotten how to do the easiest problems (of course)...so a guided demonstration (or hints) of HOW to do this problem would make me extremely happy.
 HW Helper P: 1,347 Maybe I'm dense, but have you learned how to take derivatives yet?
 P: 12 Nope, haven't even gotten that far...this is a problem from literally the first chapter of my book. It's definitely making me feel really dumb.
 HW Helper P: 1,347 Calculus Easy Slope of Curve Problem Edit: Maybe do this? $$f(x) = x^3 - 8x$$ $$Vav = \frac{f(t) - f(a)}{t - a}$$ Let a = 1, t = 1.1. Find Vav. Then let a = 1 and t = 1.01. Find the new value of Vav. Then let a = 1 and t = 1.001. Find the new value of Vav. Then let a = 1 and t = 1.0001. Find the new value of Vav. By this point, you can see what the slope of the curve will be at x = 1.
 P: 12 I've learned that, but it's in a later section than the problem I'm dealing with I think. I found a similar problem in my book where the directions state: "Slope of tangent lines: For the following functions, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at the indicated point." The problem is: f(x)=x3-x at x=1 Is this the same kind of problem I have above? The problem is not worked out in my book.
 HW Helper P: 1,347 See my previous post. (You posted while I edited, sorry!)
 P: 12 So, for the first one: Vav= (1.01)^3-8(1.01) - 1 / 1.01 - 1 ? ....that comes out to -804.97 though. Ahh, I'm sorry. I just am having a lot of trouble understanding this for some reason. Thanks so much for your help!
HW Helper
P: 1,347
No, the first iteration is actually this line (bolded):
 Quote by eumyang Edit: Maybe do this? Let a = 1, t = 1.1. Find Vav. Then let a = 1 and t = 1.01. Find the new value of Vav. ...
So
$$Vav = \frac{f(1.1) - f(1)}{1.1 - 1}$$

You did the 2nd iteration,
$$Vav = \frac{f(1.01) - f(1)}{1.01 - 1}$$

 Quote by tensirk So, for the first one: Vav= (1.01)^3-8(1.01) - 1 / 1.01 - 1 ? ....that comes out to -804.97 though.
But the bolded part is wrong. Check your math! f(1) ≠ 1.
 P: 12 I think I may have gotten it now. Plug in y= x3-8x for both f(t) and f(a) and substitute the numbers of a and t into the equations and then divide by t-a. I got numbers arbitrarily close to -5, so I believe the answer is -5. Once again thanks SO much for all of your help. :)

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