To determine the type of discontinuity in the graph of the function f(x) = x/((x+1)(x-3)), we need to look at the denominator of the function. The function will be discontinuous wherever the denominator equals zero, as division by zero is undefined.

The denominator of the function is (x+1)(x-3). Setting this equal to zero gives us the equations x+1=0 and x-3=0. Solving these equations gives us x=-1 and x=3.

Now, we need to determine the type of discontinuity at these points. There are two main types of discontinuity: removable (also known as a hole) and non-removable (which includes jumps and vertical asymptotes).

A removable discontinuity occurs when the function is undefined at a certain point, but could be defined by a slight adjustment to the function. This is not the case here, as there is no value we could assign to f(-1) or f(3) that would make the function continuous at these points.

A non-removable discontinuity occurs when there is a jump in the function or when the function approaches infinity as x approaches a certain value. In this case, as x approaches -1 or 3, the function f(x) = x/((x+1)(x-3)) approaches infinity. Therefore, the function has non-removable discontinuities at x=-1 and x=3. These are also known as vertical asymptotes.

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To determine the type of discontinuity in the graph of the function f(x) = x/((x+1)(x-3)), we need to look at the denominator of the function. The function will be discontinuous wherever the denominator equals zero, as division by zero is undefined.

The denominator of the function is (x+1)(x-3). Setting this equal to zero gives us the equations x+1=0 and x-3=0. Solving these equations gives us x=-1 and x=3.

Now, we need to determine the type of discontinuity at these points. There are two main types of discontinuity: removable (also known as a hole) and non-removable (which includes jumps and vertical asymptotes).

A removable discontinuity occurs when the function is undefined at a certain point, but could be defined by a slight adjustment to the function. This is not the case here, as there is no value we could assign to f(-1) or f(3) that would make the function continuous at these points.

A non-removable discontinuity occurs when there is a jump in the function or when the function approaches infinity as x approaches a certain value. In this case, as x approaches -1 or 3, the function f(x) = x/((x+1)(x-3)) approaches infinity. Therefore, the function has non-removable discontinuities at x=-1 and x=3. These are also known as vertical asymptotes.

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To determine the type of discontinuity in the graph of the function f(x) = x/((x+1)(x-3)), we need to look at the denominator of the function. The function will be discontinuous wherever the denominator equals zero, as division by zero is undefined.

The denominator of the function is (x+1)(x-3). Setting this equal to zero gives us the equations x+1=0 and x-3=0. Solving these equations gives us x=-1 and x=3.

Now, we need to determine the type of discontinuity at these points. There are two main types of discontinuity: removable (also known as a hole) and non-removable (which includes jumps and vertical asymptotes).

A removable discontinuity occurs when the function is undefined at a certain point, but could be defined by a slight adjustment to the function. This is not the case here, as there is no value we could assign to f(-1) or f(3) that would make the function continuous at these points.

A non-removable discontinuity occurs when there is a jump in the function or when the function approaches infinity as x approaches a certain value. In this case, as x approaches -1 or 3, the function f(x) = x/((x+1)(x-3)) approaches infinity. Therefore, the function has non-removable discontinuities at x=-1 and x=3. These are also known as vertical asymptotes.

What type of discontinuity is in the graph of f(x) = x/((x+1)(x-3))?

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