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).

2 Answers
Jul 16, 2017

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.

Jul 16, 2017

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