Question

Which of the following functions is continuous at x = 2? Select one:
a. f of x equals the quotient of the quantity x squared minus 4 and the quantity x minus 2
b. f of x equals the quotient of the quantity x squared minus 4 and the quantity x minus 2 for x not equal to 2 and equals 3 for x equals 2
c. f of x equals the quotient of the quantity x squared minus 4 and the quantity x minus 2 for x not equal to 2 and equals 4 for x equals 2
d. All are continuous at x = 2

Answer 1

Answer:

option (a) and (c).

Step-by-step explanation:

(a) The function f(x) is given as:

$f(x)=\dfrac{x^2-4}{x-2}$

which could also be represented as:

$f(x)=\dfrac{(x-2)(x+2)}{x-2}\\\\f(x)=x+2$

clearly the limit of the function at x=2 exist  (since f(x)=x+2, and limit x→2 f(x)=2+2=4)

and also f(x) is a polynomial function and hence is continuous at x=2.

(b) The function f(x) is given as:

$f(x)=\dfrac{x^2-4}{x-2}=x+2$ when x≠2 (  done in part (a))

$f(x)=3$ when x=2

for checking the continuity of a function at a point x=a we must know that:

The left hand limit(L.H.L) of f(x)(at x=a)=Right hand limit(R.H.L.) of f(x)(at x=a)=f(a)

so from here we get that L.H.L=R.H.L=4 (since f(x)=x+2,and limit x→2 f(x)=2+2=4)

but f(2)=3

Hence, the function is not continuous.

(c) The function f(x) is given as:

$f(x)=\dfrac{x^2-4}{x-2}=x+2$ when x≠2

$f(x)=4$ when x=2

Now the L.H.L=R.H.L=4 (since f(x)=x+2, and limit x→2 f(x)=2+2=4)

Also f(2)=4

Hence as L.H.L=R.H.L=f(2)

Hence, the function is continuous.

(d) As function in part (b) is not continuous.

Hence option (d) is incorrect ( All are continuous at x=2)

Hence, the answer to this question are option (a) and (c).