Are there any restrictions on the domain of $\frac { b - 3} { - b ^ { 2} + 18b - 80} \cdot \frac { b ^ { 2} - 18b + 80} { 9b }$ ?

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Answer 1 · 2017-04-07T21:56:13

$b !=0, 8, 10$
$therefore b in RR-{0, 8, 10}$

Since the denominator cannot have $0$ , we can find the restrictions on the domain in the following way.

$9b != 0$
$implies b!= 0$

$-b^2+18b-80 != 0$
$implies -b^2+10b+8b-80!=0$
$implies -b(b-10)+8(b-10)1=0$
$implies (b-10)(-b+8)!=0$
$implies b!= 10, 8$

Hence $b$ cannot be either $0$ or $8$ or $10$ .
$therefore b !=0, 8, 10$ .
$therefore b in RR-{0, 8, 10}$

Are there any restrictions on the domain of \frac { b - 3} { - b ^ { 2} + 18b - 80} \cdot \frac { b ^ { 2} - 18b + 80} { 9b }\frac { b - 3} { - b ^ { 2} + 18b - 80} \cdot \frac { b ^ { 2} - 18b + 80} { 9b }?