For the given function x=5 is not in the domain.
for h(x)=(x-6)/(x2-81), any number divided by zero is undefined.
So equating you denominator to "0" and solve for x will gives values not in the domain.
The key to finding values that are not in the domain of the function is finding any values that produce 0 in the denominator or negative numbers under radicals that have an even index .
→x²-81= x²-92=0 (difference of squares)
→(x-9)(x+9)=0
→x-9=0 and x+9=0
→x=9 and x=-9 not in the domain.
For the second question.
Again, the denominator is, so then we get x-5
→x-5=0
→x=5 is not in the domain.
To learn more about the domain functions.
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