Given: f(x) = x + 2 and h(x) = 1/ x - 1 2f(x) - h(x) = ?
2 f(x)= 2( x+2)=?
h(x) = (1/x) -1 or h(x) = 1/(x-1) ???
@jhonyy9 h(x) = 1/(x-1)
ok. - ty so can you calculi what wrote above eliesaab ?
@jhonyy9 Not really...
this will give you 2f(x) = ?
why ? f(x) = x+2 and this you need multiply it by 2 so will be 2(x+2) = ?
how you distribute this parentheses ?
@jhonyy9 would it be 2x + 4?
exactly - very nice
so and now 2f(x) -h(x) = ?
@jhonyy9 (2x + 4) - 1/ (x - 1) ?
yes and make the calculi
a-1/b = ?
how you subtract from one integer a fraction ? a-(1/b) = ?
how many will be the common denominator ?
in case of my example the common denominator will be b so this mean that you need multiply the a by b and will get (a*b -1)/b ok. ? can use this example to solve your exercise ?
what is b in case of your exercise ?
please use this example - so b in case of your exercise will be (x-1)
Would a be 2x + 4?
@jhonyy9 And 2x + 4 multiplied by x - 1 would be 2x - 4?
no 2x+4 - ( 1/(x-1) ) = ?
(2x+4)(x-1) = ?
= 2x*x +2x*(-1) +4*x +4*(-1) = ?
\[2x ^{2}+ 2x - 4\]
yes but dont forget the denominator 2f(x) -h(x) = ?
looke please in case of my example the common denominator will be b so this mean that you need multiply the a by b and will get (a*b -1)/b ok. ? can use this example to solve your exercise ?
this what you ve got above is just the part of a*b
do you understand it now ?
but you need getting (a*b -1)/b
so will be 2f(x) -h(x) = 2x+4 - (1/(x-1)) = [(2x+4)(x-1)-1]/(x-1) = ?
substitute there in place of (2x+4)(x-1) this above got value
= (2x^2 +2x-4 -1)/(x-1) = ?
-4-1 = ? this is the last step
Thank you for your help! I think I get it now :)
ok. good luck yw
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