Evaluate gh(-4)

g(x)= (4x+4)5
h(x)= -1/3 (-x+5)

I'm not getting the right answer.
would any oneeeee help me ?

xx

Can you show your work so far?

hmmm.
apparently I do it kind of differently from the standard one,
so I'm not sureee if you'll understand.
but here it goes..

g(x)= (4x+4)5
h(x)= -1/3 (-x+5)
= 1/3x -5/3

gh(x)= g(1/3x -5/3)
= 4 (1/3x - 5/3) +4 All Over 5
= 4/3x - 20/3 + 4/1 All Over 5
= 4/3x - 20/3 + 20/5 All over 5
= 4/3x -20/3 - 20 All over 5
gh (-4) = 4/3(-4) -20/3 -20 All over 5
= -16/3 -20/3 -20 All over 5
= -36/3 -20 All over 5
= -12 - 20 Over 5
= -36/5

And the answer is supposed to be
-1.6 or -8/5...

Thanks for showing your work! :)

Since you want to solve for when x=-4 you don't need to do it the hard way and pull the x all the way through. Just substitute the -4 straight away and it will be much easier.



Hope this helps!

Now, your method will work also... but you have an error. I'm not sure why you converted the 4 into a 20/5, that was OK but of no use. But for some reason you then converted the 20/5 into a 20, that is your error. If you had left it as 4 or 20/5 you would have gotten 8/5.

h(-4) = -1/3(-(-4)+5) = -1/3(9) = -3
gh(4) = g(-3) = (4(-3)+4)5 = (-8)5 = -40

g(x)=3x/x+1
Find 1) g2(x) 2)g3(x)

Hello Samantha16, it would be good if you started your own thread.



Find
1)
2)



Can you continue now? :)

Do the same thing, but a little further for the second one. Post what you get! :)

Hello, umm I've got an alevel question and wanted to know what the answer is but but can't find the mark scheme;

f(x) = 2e(power x) domain minus infin. To infin.

g(x) = 3lnx domain [1, infin.)

a) Explain why gf(-1) does not exist?
b) Find its simplest form an expression for fg(X). State the domain and range of fg.

fg(X) = 2e(to the power of)3lnx
= 2(3x)
= 6x

but don't know the domain and range or part a!


Thanks :)

You could have posted a new thread for this...

Anay, I wanted to confirm the question.





And a) asks to show that does not exist?

And b) asks to simplify fg(x)

If so...

a) Does exist it seems...



Which is a straight line... in case it's the inverse, that too would exist.

b) You didn't do it correctly, I'm afraid.



And since this function usually exists from -infinity to +infinity, you have to take the smallest domain that you were given, that is [1, + infinity)

The range is then calculated. What is fg(x) at x = 1. From there, you get your lowest value, until infinity. You range is thus from that value to positive infinity.