Question

Discuss the continuity of the following function at the point(s) or in the interval indicated against them:

f(x) = 2x² − 2x + 5  for 0 ≤ x < 2

= `(1 - 3x - x^2)/(1 - x)`     for 2 ≤ x < 4

= `(7 - x^2)/(x - 5)`  for 4 ≤ x ≤ 7 on its domain.

Answer 1

The domain of f is [(0, 5) ∪ (5, 7)]
We observe that x = 5 is not included in the domain as f is not defined at x = 5
a. For 0 ≤ x < 2
f(x) = 2x² – 2x + 5
It is a polynomial function and is continuous at all point in [0, 2)]

b. For 2 < x < 4

f(x) = `(1 - 3x - x^2)/(1 - x)`

It is a rational function and is continuous everwhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 1
But x = 1 does not lie in the interval.
f(x) is continuous at all points in (2, 4)

c. For 4 < x ≤ 7, x ≠ 5 i.e. for x ∈ [(4, 5) ∪ (5, 7)]

f(x) = `(7 - x^2)/(x - 5)`

It is a rational function and is continuous everwhere except possibly at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 ∉ [(4, 5) ∪ (5, 7)]
∴ f is continuous at all points in [(4, 7] – {5}.

d. Since the definition of function changes around x = 2, x = 4 and x = 7
∴ there is disturbance in behaviour of the function. So we examine continuity at x = 2, 4, 7 separately.
Continuity at x = 2 :

`lim_(x→2^-) "f"(x) = lim_(x→2^-) (2x^2 – 2x + 5)`

= 2(2)² – 2(2) + 5
= 8 – 4 + 5
= 9

`lim_(x→2^+) "f"(x) = lim_(x→2^+) (1 - 3x - x^2)/(1 - x)`

= `(1 - 3(2) - (2)^2)/(1 - 2)`

= `(1 - 6 - 4)/-1`

= `(-9)/(-1)`

= 9

Also, f(x) = `(1 - 3x - x^2)/(1 - x)`, at x = 2

∴ f(2) = `(1 - 3(2) - (2)^2)/(1 - 2)`

f(2) = 9

∴ `lim_(x→2^-) "f"(x) = lim_(x→2^+) "f"(x)` = f(2)

∴ f is continuous at x = 2

e. Continuity at x = 4:

`lim_(x→4^-) "f"(x) = lim_(x→4^-) (1 - 3x - x^2)/(1 - x)`

= `(1 - 3(4) - (4)^2)/(1 - 4)`

= `(1 -12 - 16)/(1 - 4)`

= `(-27)/(-3)`

= 9

`lim_(x→4^+) "f"(x) = lim_(x→4^+) (7 - x^2)/(x - 5)`

= `(7 - (4)^2)/(4 - 5)`

= `(7 - 16)/-1`

= 9

f(x) = `(7 - x^2)/(x - 5)`, at x = 4

∴ f(4) = `(7 - (4)^2)/(4 - 5)`

= 9

∴ `lim_(x→4^-) "f"(x) = lim_(x→4^+) "f"(x)` = f(4)

∴ f is continuous at x = 4