Let f be given by the formula?

Question

Let f be given by the formula:

Find the value(s) of x at which f is discontinuous?

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Answer 1 · 2018-03-01T17:26:11

At #x=1#

Consider the denominator.
#x^2 + 2x -3#
Can be written as:
#x^2 + 2x +1 -4#
#(x + 1)^2 -4#
#(x + 1)^2 -2^2#

Now from relation #a^2-b^2# = #(a+b)(a-b)# we have
#(x+1 +2)(x+1 -2))#

#(x+3)(x-1))#

If #x=1#, the denominator in above function is zero and the function tends to #oo# and not differentiable. Is discontinous.

Answer 2 · 2018-03-01T17:46:35

#f(x)=(x+2)/(x^2+2x-3)# is discontinuous when #x=-3# and #x=1#

#f(x)=(x+2)/(x^2+2x-3)# is discontinuous when denominator is zero i.e.

#x^2+2x-3=0#

or #x^2+3x-x-3=0#

or #x(x+3)-1(x+3)=0#

or #(x-1)(x+3)=0#

i.e. #x=-3# and #x=1#

graph{(x+2)/(x^2+2x-3) [-10, 10, -5, 5]}