Is this correct? p=10 q=110 r=70
2 answers:
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To me personally, the first bit f(g(x)) is easy and the domain is tricky. Let's try explain this.
A function takes an input number and returns an output number depending on the function. Look at f(x) = x+3, if we let the input number be 2 then we say that f(2) = 5. We could do f(π) to give us π+3 or even f(x²) to give us x² +3. The trick is to substitute the input into the function equation.
You have been asked to find f(g(x)). You know f(x) =
. Putting numbers in at this point would be easy (try work out f(2), you'll do it really quick) but you have to put in g(x).
f(g(x)) =
we also know that g(x) =
so we can say that
f(g(x)) =
and that is f(g(x)) but the question requires that we simplify it so
f(g(x)) =
Now for the tricky bit (for me, at least). The domain is the full set of values that you can 'put in to' the function and still get a real value out. So how do we work out what numbers 'break' the function? I like to use the fact that DIVIDING BY ZERO IS IMPOSSIBLE. What value of x can we put into the function to make it so the function is being divided by 0? i.e. 1-2x = 0 solve that and you have a value of x that isn't part of the domain.
This means the domain is all real numbers EXCEPT the solution to that equation. (Because if we put that value into f(g(x)) it's impossible to get a value out.)
[I know this was a lot to read, if you have any questions or don't get anything feel free to message me or leave a comment.]
A function takes an input number and returns an output number depending on the function. Look at f(x) = x+3, if we let the input number be 2 then we say that f(2) = 5. We could do f(π) to give us π+3 or even f(x²) to give us x² +3. The trick is to substitute the input into the function equation.
You have been asked to find f(g(x)). You know f(x) =
f(g(x)) =
we also know that g(x) =
f(g(x)) =
f(g(x)) =
Now for the tricky bit (for me, at least). The domain is the full set of values that you can 'put in to' the function and still get a real value out. So how do we work out what numbers 'break' the function? I like to use the fact that DIVIDING BY ZERO IS IMPOSSIBLE. What value of x can we put into the function to make it so the function is being divided by 0? i.e. 1-2x = 0 solve that and you have a value of x that isn't part of the domain.
This means the domain is all real numbers EXCEPT the solution to that equation. (Because if we put that value into f(g(x)) it's impossible to get a value out.)
[I know this was a lot to read, if you have any questions or don't get anything feel free to message me or leave a comment.]
Answer: 9. x=-3/2 10. x=2
x= 5/2 x=5/3
Step-by-step explanation:
9. Factor the equation:
How: put 4x^2 in the bottom left box and -15 in the upper right box.
then multiply 4x^2 x -15 = -60x^2 and put your answer in the top center box of the diamond.
Put the -4x from the equation in the bottom box. Then list out factors that multiply to the top box (-60x^2) and add to the bottom box (-4x) The two factors are -10 and 6
-10 x 6 = -60
-10 + 6 = -4
Put -12x and -5x in the remaining white boxes and find what numbers multiply to each.
I attached a picture. The colored dots correspond with each other.
Eg: Orange dot x orange dot = number in white square
Now you get the two factors (2x+3)(2x-5)
Set both of these equal to 0 and solve
0=2x+3 0=2x-5
x= -3/2 x= 5/2
Do the same for the other problem
