Identify the functions that exhibit a removable discontinuity.

Answer:

h(x) and k(x)

Step-by-step explanation:

The functions that exhibit a removable discontinuity are function i(x) and function k(x)

"An equation of the form ax² + bx + c = 0 is a quadratic equation."

"It defines a relation between input and output values."

"This function has a discontinuity at one or more values mainly because of the denominator of a function is being zero at that."

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