# computing the slope of the secant line between points

#### JDS

Hello everyone. I am really getting lost trying to figure out how to correctly figure this one out. Here is the problem:

Compute the slope of the secant line between the points at (a) x = 1 and x = 2, (b) x = 2 and x = 3, (c) x = 1.5 and x = 2, and (d) x = 2 and x = 2.5, (e) x = 1.9 and x = 2, (f) x = 2 and x = 2.1, and (g) use parts (a) - (f) and other calculations as needed to estimate the slope of the tangent line at x = 2.....

f(x) = ex

To give some context here, I have been learning (In Calculus: Early Transcendental Functions) about Differentiation, and specifically tangent lines and velocity.

Any help with this problem would be greatly appreciated.

#### hollywood

The point of the problem is that as the second point approaches x=2 (from both directions), the secant line approaches the tangent line.

What is a secant line? You just take two points on the graph and draw a line through them. So in (a), for example, you calculate f(1) and f(2) which are your y-values, and then just calculate the slope as:

$$\displaystyle \frac{y_2-y_1}{x_2-x_1}$$

- Hollywood

#### JDS

Thanks Hollywood, I understand it a lot better now, however I do not quite understand how I am supposed to calculate f(1) and f(2), for example in (a), because the problem has f(x) = ex , So what am I supposed to do with the "e" or have I missed something somewhere?

From what I gathered, to find f(x) = ex , replace the x with 1,which gives us f(x) = e1 , so what now?

#### hollywood

The symbol e just represents a number $$\displaystyle e=2.71828...$$. You probably have the $$\displaystyle e^x$$ function on your calculator.

So $$\displaystyle e^1=e$$ is just a number, and so is $$\displaystyle e^2$$, $$\displaystyle e^{1.5}$$, etc.

So for part (a), for example, the slope is:

$$\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{e^2-e^1}{2-1}=e^2-e\approx{4.67}$$

- Hollywood

#### JDS

Got ya, thanks Hollywood!!!