About first derivative test (ID test)

Question

Q. Find the intervals on which f is increasing and decreasing.

f(x)= (x^2-24)/(x-5)

-> The answer is that f is increasing in (negative infinity, 4), (6, positive infinity) and decreasing in (4, 5), (5, 6).

When we evaluate where does the f is increasing or decreasing, we have to find the critical points that make f'(x)=0 or f'(x) undefine. I've known that if the critical points that I found are not included in the function's domain, then we couldn't say that point is a critical point. In this problem, I can see that the point x=5, where f'(x) is undefined at this point, is not in our domain of f (because the domain of f is x≠5). How do we use 5 in our I/D test even though it isn't included in our domain?

Answer 1

(x^2-24)/(x-5)

= (x^2-24)(x-5)^-1

f' = (x^2-24)(-1)/(x-5)^2 +(2x)/(x-5)

= [24-x^2 +2x^2-10x]/(x-5)^2

=(x^2 -10x +24)/(x-5)^2 =0

x^2 +10x+24 = 0

(x-4)(x-6) = 0

x=4,6, points where slope changes direction

at 4 and 6 slope = 0

at 5 slope is undefined

or just use a graphing calculator and read off 4 and 5 as the places where the slope changes direction from + to -. intervals (-inf,4), (6,inf) have increasing slope

x=5 is a discontinuity in the graph with the intervals (4,5) and (5,6) both decreasing slope

f" =((x-5)(2x-10) - (x^2-10x+24)(2x-10))/(x-5)^4

=(2x^2-20x+50 -2x^3+30x^2-148x+240))/(x-5)^4

= 2/(x-5)^3

at less than 5, f"<0 and concave down

at >5, f">0 and concave up

critical POINTS are POINTS. x=5 is not a point, there is no point when x=5

critical points include relative extrema and inflection points

an inflection point also has f"=0, unlike a discontinuity which has no derivatives, no f' or f" defined

check your math calculations by graphing the function, with an online graphing calculator such as Desmos

Answer 2

This function has critical points at x=4 and x=6, 4 is a max and 6 is a min and x=5 it has a discontinuity (i.e. as you say the point x=5 is not in the domain).

You get the critical points in the usual way, by setting the derivative = 0.

In my (not so humble) opinion the easiest way to find the increasing and decreasing intervals is from a graph!