Identify the functions that exhibit a removable discontinuity.

"If adjusting the function's value at the point of discontinuity will render the function continuous, then such discontinuity is called removable discontinuity."

What is non-removable discontinuity?

"If adjusting the function's value at the point of discontinuity will not render the function continuous, then such discontinuity is called non-removable discontinuity."

For given question,

1) Consider a function [tex]f(x)=\frac{x^{2} +4x-21}{x^{2} -13x+42}[/tex]

If we factorize the quadratic equation at numerator and denominator then the given function would be,

[tex]f(x)=\frac{(x-3)(x+7)}{(x-7)(x-6)}[/tex]

We can observe that the function f(x) is discontinuous at x = 7 and x= 6.

So, the function f(x) has non-removable discontinuity.

2) Consider a function [tex]g(x)=\frac{2x^{2} -5x+12}{2x+3}[/tex]

We can observe that the function g(x) is discontinuous at x -3/2.

We cannot factorize the quadratic equation at numerator.

So, the function g(x) has non-removable discontinuity.

3) Consider a function [tex]h(x)=\frac{x-3}{x^{2} -9}[/tex]

If we factorize the quadratic equation at denominator then the given function h(x) would be,

[tex]\Rightarrow h(x)=\frac{(x-3)}{(x-3)(x+3)}\\\\\Rightarrow h(x)=\frac{1}{x+3}[/tex]

We can observe that the function h(x) is discontinuous at x = -3

So, the function h(x) has non-removable discontinuity.

4) Consider a function [tex]i(x)=\frac{x^{2} -8x+15}{x-5}[/tex]

If we factorize the quadratic equation at numerator then the given function would be,

[tex]\Rightarrow i(x)=\frac{(x-5)(x-3)}{x-5}\\\\ \Rightarrow i(x)=x-3[/tex]

We can observe that the function i(x) is a continuous function.

So, the function i(x) has removable discontinuity.

5) Consider a function [tex]j(x)=\frac{x+4}{-2x+8}[/tex]

We can observe that the function j(x) is discontinuous at x = 4

And the function j(x) has non-removable discontinuity.

6) Consider a function [tex]k(x)=\frac{2x^{2} +x-15}{2x-5}[/tex]

If we factorize the quadratic equation at numerator then the given function would be,

[tex]\Rightarrow k(x)=\frac{(x+3)(2x-5)}{(2x-5)}\\\\ \Rightarrow k(x)=x+3x^{2}[/tex]

We can observe that the function k(x) is a continuous function.

So, the function k(x) has removable discontinuity.

Therefore, the functions that exhibit a removable discontinuity are function i(x) and function k(x)

Learn more about the removable discontinuity here:

https://brainly.com/question/9648611

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