Find the dicontinuities of the function. f(x) = x2 + 12x + 27 x2 + 4x + 3 . There is a removable discontinuity at (, ).
2 answers:
Solution-
Equating the denominator to 0,
⇒ (x+3)(x+1) = 0
⇒ x= -3, -1
∴ Discontinuities of the function f(x) is at -3, -1
As (x+3) is also a factor of the numerator, you will have what's called a removable discontinuity. That will show up as a hole in the graph.
∴ At x = -3, you will have removable discontinuity.
You might be interested in
Rewrite the limand as
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = (1 - sin(<em>x</em>)) / (cos²(<em>x</em>) / sin²(<em>x</em>))
… = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / cos²(<em>x</em>)
Recall the Pythagorean identity,
sin²(<em>x</em>) + cos²(<em>x</em>) = 1
Then
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / (1 - sin²(<em>x</em>))
Factorize the denominator; it's a difference of squares, so
1 - sin²(<em>x</em>) = (1 - sin(<em>x</em>)) (1 + sin(<em>x</em>))
Cancel the common factor of 1 - sin(<em>x</em>) in the numerator and denominator:
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = sin²(<em>x</em>) / (1 + sin(<em>x</em>))
Now the limand is continuous at <em>x</em> = <em>π</em>/2, so