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- Thread starter Zayne
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Unsure on 5 and first parts of 7.
For 7a just put the stated x values in the function and those will be the values of b(x).
For 7c you'd make b(x) = x and x = b'(x) and re-arrange to make b'(x) the subject. I got b'(x) = 2 + x / x - 2 . Range would be when b'(x) <= -3 and b'(x) >= 5 & domain is x cannot be 2.
For 7d , b(1) would mean denominator => 1-1 = 0 which has no solutions because you can't divide by zero and likewise with b'(2) => 2-2 = 0.
Believe these are correct but I'm not that great at maths.
For 7a just put the stated x values in the function and those will be the values of b(x).
For 7c you'd make b(x) = x and x = b'(x) and re-arrange to make b'(x) the subject. I got b'(x) = 2 + x / x - 2 . Range would be when b'(x) <= -3 and b'(x) >= 5 & domain is x cannot be 2.
For 7d , b(1) would mean denominator => 1-1 = 0 which has no solutions because you can't divide by zero and likewise with b'(2) => 2-2 = 0.
Believe these are correct but I'm not that great at maths.
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- #4
You can try doing the question yourself doing what Piz said. Here is my take on q7, I'll attempt to translate what q5 wants cause me no English no no no.
The range and domain switch places for the inverse relation. It's simple to find the domain G for b(x) at the start but a bit more tricky to find range H, that's why I skipped finding H till I found b^-1(x). Since H is it's domain, it's easy to find.
For d) you just explain that you can't divide by 0
The range and domain switch places for the inverse relation. It's simple to find the domain G for b(x) at the start but a bit more tricky to find range H, that's why I skipped finding H till I found b^-1(x). Since H is it's domain, it's easy to find.
For d) you just explain that you can't divide by 0
