Find the holes
itle="f(x)=\frac{x+3}{(x+1)(x-2)}" alt="f(x)=\frac{x+3}{(x+1)(x-2)}" align="absmiddle" class="latex-formula">
1 answer:
Answer:
<h2>
x=-1 and x=2 </h2>
Step-by-step explanation:
The "holes" are the x-es where function doesn't exist.
one must not divide by 0, so given function doesn't exis if the denominator is equal 0
The product {(x+1)×(x-2)} is equal 0 if any of factors {(x+1),(x-2)} is 0, so
x + 1 = 0 or x - 2 = 0
x = - 1 or x = 2
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Step-by-step explanation:
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