Question

If `y=x^2+3x` then find the slope of the secant line at `x=3` and `x=3+h`. What happens as `Deltax rarr 0`? Take `lim_(Deltax->0) (f(x+Deltax) - f(x))/(Deltax)`.

Answer 1

Well, I assume you know how to find a slope in general, just not how to apply it here:

`"Slope" = (Deltaf(x))/(Deltax) = (f(x_f) - f(x_i))/(x_f - x_i)`

If you want to estimate the slope of `f(x) = x^2 + 3x` at `x = 3`, you can pick points `(x,f(x))` around/near `x = 3` so that you can find the slope of a line that touches `f(x)` at `x = 3`. The closer we are to `x = 3` with our chosen points, the more accurate we will be.

Say we chose `x_f = 3.1` and `x_i = 2.9`. This gives:

`f(3.1) = (3.1)^2 + 3(3.1) = 18.91 = f(x_f)`

`f(2.9) = (2.9)^2 + 3(2.9) = 17.11 = f(x_i)`

So, the slope around `x = 3` with a spread of `Deltax = 3.1 - 2.9 = 0.2` gives:

`color(blue)("Slope") ~~ (18.91 - 17.11)/(3.1 - 2.9)`

`~~ color(blue)(9)`

graph{(y - x^2 - 3x)(y - 18 - 9(x - 3)) = 0 [-10, 10, -9.65, 48.9]}

The above graph shows the straight line that represents the slope at `x = 3`.

The true slope at `x = 3` is found by taking the derivative, which is another way of saying "let `Deltax -> 0` and let's see what the slope becomes".

`lim_(Deltax->0) (f(x_f) - f(x_i))/(x_f - x_i)`

`= (f(x_i + Deltax) - f(x_i))/(x_f - x_i)`

A simple way to take a derivative, `(df)/(dx)`, of a polynomial is with the power rule:

`d/(dx)[x^n] = nx^(n-1)`

So, for `f(x) = x^2 + 3x`:

`(df)/(dx) = 2 cdot x^(2-1) + 3(1 cdot x^(1-1))`

`= 2x + 3`

And at `x = 3`, the derivative gives the slope at `x = 3`:

`color(blue)(|[(df)/(dx)]|_(x=3)) = 2(3) + 3 = color(blue)(9)`

So, it turns out that we chose our `Deltax` in such a way that our estimated, average slope was the actual slope.

Answer 2

Slope = `9`

Explanation:

I will answer this question with reference to a similar relevant question that you asked:

In this similar question:

https://socratic.org/questions/show-that-the-expression-for-the-slope-of-the-secant-line-through-y-x-2-3x-at-x-#451989

we were asked to show that the expression for the slope of the secant line through `y = x^2+3x` at `x =3` and `x = 3+h` is:

`m_(sec) = h+9`

So if we take small values of `h` (eg `h=0.01`) then we get a reasonable estimate of the slope of the tangent at `x=3`, in this case, our estimate would be slope`=0.01+9=9.01`.

In this question, I further demonstrated that as `h rarr 0` then the secant line approaches the tangent with more and more accuracy, and, so in the limit we have:

`m_(tan) = lim_(h rarr 0) m_(sec)`

`" " = lim_(h rarr 0) (h+9)`

`" " = 9`