# Calculus Easy Slope of Curve Problem

• tensirk

## Homework Statement

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.

## Homework Equations

Would it be...
Vav= s(t)-s(a)/t-a?

## 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.

Maybe I'm dense, but have you learned how to take derivatives yet?

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.

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.

Last edited:
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.

See my previous post. (You posted while I edited, sorry!)

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!

No, the first iteration is actually this line (bolded):
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,
$$Vav = \frac{f(1.01) - f(1)}{1.01 - 1}$$

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.

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. :)