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junior college 1 | H2 Maths
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Need help for part iii please. Thanks
Date Posted: 4 years ago
Views: 245
ln(x² - x)
= ln(x² - 2x + 1 + x - 1)
= ln( (x - 1)² + (x - 1) )
= ln ( (1/f(x))² + 1/f(x) )
( Since
f(x) = 1/(x - 1) ,then 1/f(x) = (x - 1)
(f(x))² = 1/(x - 1)² , then 1/(f(x))² = (x - 1)² )
so g(x) = ln (1/x² + 1/x)
= ln(x² - 2x + 1 + x - 1)
= ln( (x - 1)² + (x - 1) )
= ln ( (1/f(x))² + 1/f(x) )
( Since
f(x) = 1/(x - 1) ,then 1/f(x) = (x - 1)
(f(x))² = 1/(x - 1)² , then 1/(f(x))² = (x - 1)² )
so g(x) = ln (1/x² + 1/x)
Alternative to completing the square :
ln(x² - x)
= ln( x(x-1) )
= ln ( (x - 1 + 1)(x - 1) )
= ln ( (x - 1)(x - 1) + (x - 1) )
= ln ( (x - 1)² + (x - 1) )
Continue as before
ln(x² - x)
= ln( x(x-1) )
= ln ( (x - 1 + 1)(x - 1) )
= ln ( (x - 1)(x - 1) + (x - 1) )
= ln ( (x - 1)² + (x - 1) )
Continue as before
See 1 Answer
done
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clear
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Didn’t see J’s comment but if you don’t see the trick for completing the square just simplify as much as you can and then work backwards so to speak. Ask yourself what g should be to give the gf given f. You can see that the g should be ln(x) where x can be manipulated to give you correct form. Actually quite trial and error if you can’t see the trick but still can be done.
Date Posted:
4 years ago
Thank you :)
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