Discontinuities

Question

f(x) = [sin(x²-4)]^-1 , how do I prove whether the function has any discontinuities or not?

Answer 1

if you mean [sin(x^2-4)]^-1 =1/sin(x^2-4) = csc(x^2-4) = y then graphically y has a gap or discontinuity, as y cannot be between -1 and +1. Just graph it, use an online graphing calculator

cosecants of angles are the hypotenuse of a right triangle divided by the opposite side's length, wich can be negative. By definition the hypotenuse is the largest side.

but also other discontinuities where x=1 or -1, then sin(x^2-1) = sin(1^1-1) = sin(0) = 0, as that makes the denominator of 1/sin(x^2-4) undefined.

BUT if you meant to write sin^-1(x^2-4)

then

f(x) values are the angles whose sine = x^2-4

let x^2-4 = y

-1 < y < 1 or -1 is less than x^2-4 is less than +1

sines of angles cannot be less than -1 or greater than +1

sines are the opposite side over the hypotenuse of a right triangle, and the hypotenuse is always greater than either of the triangle's other sides

try using a graphing calculator such as Desmos on line. It shows two separate lines, with a large gap/discontinuity between them. but notice the y values cover all values from -sqr3 to +sqr3, with no gap or disconinuity

or try graphing x^2-1. It's a parabola with vertex -4, and x intercepts x=2 and x-2, You can, by hand, draw a rough sketch

but you want only the points above the parabola and also between the horizontal lines: y=1 and y=-1

and the points between the vertical lines x=-sqr5 and x=-sqr3, and the points between x=sqr3 and x=sqr5

use a graphical calculator, you'll get two separate parts, with a discontinuity, a gap between them from x=-sqr3 to x=sqr3

try any point inbetween, such as x=0, the angle doesn't exist as no angle has a sine of -4

try x=2 and x=-2, then the sine of the angle =0 and the angle = 0 or pi or 180 degrees plus 2npi or 360n where n= any integer. but by convention, the range of sin^-1 or also known as arcsin is limited to [-pi/2,pi/2]

the angles are not discontinuous, as they span the full range from -pi/2 to pi/2 or from -90 to +90 degrees

yet the points on the graph are discontinuous for the x values, as the x values are only [-sqr5, sqr3] U [-sqr5,-sqr3] and nowhere inbetween those two discontinuous sets

Look at the graph and the y values are not discontinuous, as they span the full range from -1 to +1, and the y values' continuity shows/proves the continuity of f(x), the angles

or solve the inequality -1 < x^2 -4 < 1

add 4 to both sides and the middle,

then take the square roots

-sqr5 < x < -sqr3 and sqr3 < x < sqr5

it's about

-2.2 < x < -1.7 and 1.7 < x < 2.2

which is discontinuous, but x^2-1 and f(x) aren't discontinuous