Answer
(a) Discontinuity of the function occurs when the denominator term becomes zero
$\mid x\mid -3 = 0.$
At x=3 and x=-3 the above function becomes zero
It is continuous on the interval $ (-\infty,-3) U (-3,3) U(3,\infty)$
(b)
$cos(1/x)$ is continuous everywhere except at x=0
$cos^{-1}(1/x)$ is continuous everywhere on the range of $cos(1/x)$ $[-1,1]$
(c)
The denominator is zero when $2x^2+3x-2 = 0$
The roots of the above equation are $(2x-1)(x+2) = 0$
Therefore the function is continuous at $(-\infty,-2)U(-2,1/2)U(1/2,\infty)$
Work Step by Step
(a) Discontinuity of the function occurs when the denominator term becomes zero
$\mid x\mid -3 = 0.$
At x=3 and x=-3 the above function becomes zero
It is continuous on the interval $ (-\infty,-3) U (-3,3) U(3,\infty)$
(b)
$cos(1/x)$ is continuous everywhere except at x=0
$cos^{-1}(1/x)$ is continuous everywhere on the range of $cos(1/x)$ $[-1,1]$
(c)
The denominator is zero when $2x^2+3x-2 = 0$
The roots of the above equation are $(2x-1)(x+2) = 0$
Therefore the function is continuous at $(-\infty,-2)U(-2,1/2)U(1/2,\infty)$
