Identify all singular points and points of discontinuity of the given function.

Question

20-5x if x≤4

f(x)=x+1 if 4<x<6

x^2-29 if x≥ 6

singular points x=

points of discontinuity x=

Answer 1

There is three conditions must be met for the function to be continuous at a point a

1. f(a) is defined
2. lim x→a f(x) exists
3. lim x→a f(x) = f(a)

let us check these conditions for x=4

f(4) = 20 - 5*4 =0

lim x→4^- f(x) = lim x→4^- (20-5x)= 20-5*4=0

lim x→4⁺ f(x) = lim x→4⁺ (x+1) = 4+1= 5

since lim x→4^- f(x) ≠ lim x→4⁺ f(x) , therefore the limit does not exist at x=4 and we have discontinuity.

check these conditions also for x=6

f(6) =36 - 29 =7

lim x→6^- f(x) = lim x→6^- (x+1)= 6+1=7

lim x→6⁺ f(x) = lim x→6⁺ (x²- 29 ) = 36-29 = 7

since lim x→6^- f(x) = lim x→6⁺ f(x) ,

therefore the limit at x=6 is exist, lim x→6 f(x) = 7, so all the conditions is met and we have continuity at x=6, but x=6 is singular point when you check for derivative for x<6 and x≥6 , you will get different values. Try to graph the function and you can see that clearly.

Answer 2

The only candidates for singularities or discontinuities are

x=4 and x=6, because everywhere else the function is continuous and differentiable.

x = 4 is a discontinuity, because its limit there is undefined.

x = 6 is a singular point, because f(x) is continuous at x = 6, but

its derivative from the left is 1 and its derivative from the right is 12.