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
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
Say we chose
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
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
The true slope at
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,
d/(dx)[x^n] = nx^(n-1)
So, for
(df)/(dx) = 2 cdot x^(2-1) + 3(1 cdot x^(1-1))
= 2x + 3
And at
color(blue)(|[(df)/(dx)]|_(x=3)) = 2(3) + 3 = color(blue)(9)
So, it turns out that we chose our
Slope =
Explanation:
I will answer this question with reference to a similar relevant question that you asked:
In this similar question:
we were asked to show that the expression for the slope of the secant line through
m_(sec) = h+9
So if we take small values of
In this question, I further demonstrated that as
m_(tan) = lim_(h rarr 0) m_(sec)
" " = lim_(h rarr 0) (h+9)
" " = 9