7 replies to this topic

[Rocky]

If you can help, it'd be really appreciated. Thanx.

1- 2 Functions are defined on suitable domains and are given as f(x)= x + 3 & g(x)= x squared - 1
(i) Find an expression for the function H where H(X)= g( f(x) )
(ii) Find the values of A given that H(a)= f(A) + 1

2- On a suitable set of REAL numbers functions F and G are defined by
f(x)= 1/x + 2 **this on left all a fraction** AND g(x)= 1/x (- 2) **1st part of it is fraction the minus 2 isn't in the fraction**
Find f( g(x) ) in its simplest form.

[dfx]

1. (i)

For h(x) = g(f(x):

you first take g(x) = x - 1

then g(f) = f - 1

then g(f(x) = f(x) - 1

= (x+3) - 1

h(x)=g(f(x)=x + 6x + 8 h(x) = x + 6x + 8

Not too sure about (ii), I'll get back to you on it.

2.

Exactly same method

For f(g(x)):

f(x) = 1/(x+2)

Then f(g) = 1/(g+2)

Then f(g(x)) = 1/(g(x) + 2)

= 1/[(1/x)-2] <- you forgot to put the + 2 back in this line [edit from Dave]

=1/[(1/x) - (2x/x)

=1/(1-2x)/x

=x/(1-2x)

^ I've done this in a hurry so someone check the algebra please. Cheers.

[Dave]

H(X) = g(f(x))

right so the function g(x) = x² - 1

so if we sub f(x) for the x in the function above we get

g(f(x)) = (x+3)² - 1 = H(X)

ii)

the value inside the brackets there the "a" is just the variable in the function

so we get

H(a) = (a+3)² - 1 = (a+3) + 1
a² +6a + 9 - 1 = a + 4
a² + 5a + 6 = 0
(a+6)(a-1) = 0
a = -6 and a = 1

2)

f(g(x)) so thats the function of f where all the variable we sub in the function g(x)

= 1/ (1/x(-2)) +2) = x -1

i think

NB: dfx if you look at my wee correction there and get back to me

[dondon]

i)

its hard to explain but

H(x)=g(f(x))
=g(x+3)
= (x+3) -1
= x +6x+8

so you put in the f(x) function then put that into the g(x) fuction, but put it into each place where there is an x it's easier to see what to do if you see it more like:
g(x)=x[^2] -1

then put in x+3 where there is an x.

ii)
i think that you do this

H(a)=f(A)+1
=(a+3)+1
=a+4

but i could be wrong

2)
i don't know if i'm doing that one wrong but it comes out with x/0 which doesn't exist but you should just be able to do the same thing as the first question but its slightly more comlicated. And remember when you divide a fraction by a fraction you turn one upside down and multiply them, thats where i think i went wrong.

[Dave]

dondon for part to you are looking to find the value of a

i think wot i have done is correct but please look and see and tell me if its not (it happens a lot lol)

[dondon]

yip i didn't realise that, you can tell my brain is really slowing down

[Rocky]

Thanx dave, dfx and dondon for the help.

[dfx]

You're welcome.

Error well spotted Dave, cheers.