To find the removable discontinuities, we need to look for values of x that make the denominator of the function equal to zero.
1.) y = (5(x-1)(x+2))/((x-5)(x+2))
The function has a removable discontinuity at x = -2 because the denominator (x+2) becomes zero. It is removable because the (x+2) in the numerator cancels out with the (x+2) in the denominator.
2.) y = ((x-3)(x+6))/((x+2)(x-3))
The function has a removable discontinuity at x = 3 because the denominator (x-3) becomes zero. The (x-3) term in the numerator cancels out with the (x-3) in the denominator, making it removable.
3.) y = (3x(x+2))/((x-5)(x+2))
The function has a removable discontinuity at x = -2 because the denominator (x+2) becomes zero. The (x+2) term in the numerator cancels out with the (x+2) in the denominator, making it removable. It also has a removable discontinuity at x = 5 because the denominator (x-5) becomes zero. The (x-5) term in the numerator cancels out with the (x-5) in the denominator, making it removable.
Note: A removable discontinuity occurs when there is a factor in the denominator that also exists as a factor in the numerator and cancels out, resulting in a hole in the graph of the function.
what is the removable DISCONTINUITIES of each:
1.) y=\frac{5\left(x-1\right)\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}
2.) y=\frac{\left(x-3\right)\left(x+6\right)}{\left(x+2\right)\left(x-3\right)}
3.)y=\frac{3x\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}
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