Question

Two functions, f and g are defined as f(x)=xsquared -2 and g(x)=2x-3.
a) 1)Calculate the value of g(-3)
2)calculate the value of x for which g(x)=3
3)calculate the value of x for which the
function f(x)=g(x)
B) Write expressions for:
1) fg(x) 2) f -1(x)..(f inverse of x) 3) (gf)-1 (x)..(g of f inverse of x)

Answer 1

Do you understand how to solve part a) 1) ? :)

Answer 2

yes i do understand how to solve part a) 1

Answer 3

g(x)=2x-3
g(-3)=2(-3) -3
=-6-3
=-9
therefore, the value of g(-3) is -9

Answer 4

Ok good :)

\(a)2)\) Calculate the value for which g(x)=3.
We know that \(g(x)=2x-3\).
But now they're asking us, when does \(g(x)\) also equal \(3\).

So we can equate these things, since they both equal \(g(x)\).\[\large g(x)=2x-3\]\[\large g(x)=3\]\[\large 3=2x-3\]Then from there, just solve for x!

Answer 5

so the answer to that would be
g(x)=2x-3
g(x)=3
therefore 3=2x-3
3+3=2x
6/2=2x/2
3=x
so the value of x for which g(x)=3 is 3

Answer 6

So g(x)=3, when x=3. Ok good :)

Answer 7

\(\large a)3)\)
Ok this one is a little tricky. We want to know when the functions are equal. What this means is ~ if you were to graph the functions, where would the lines intersect?
To solve this part, we set them equal to each other, and solve for x.\[\large \begin{align*}f(x)&=g(x)\\x^2-2&=2x-3\end{align*}\]

See how that equation on the left is our f(x)?
From here, we want to get everything on one side, and solve this quadratic.

Answer 8

Okay the answer would be
f(x)=g(x)
x2-2=2x-3
x2-2x +3-2=0
(X-1) (X+2)=0
X-1=0
X=-1 and x+2=0
x=-2
so, based on the quadractic principle where 'x' has two values, the values for 'x' are 1 and -2

Answer 9

\[\large x^2-2x+1=0\]I think this actually factors into,\[\large (x-1)(x-1)=0\]

Answer 10

yes the factors are correct but where did the +1 come from and which question is this
?

Answer 11

a) 3)
You were able to get to this step,\[\large x^2-2x +3-2=0\]Which simplifies to,\[\large x^2-2x+1=0\]

Answer 12

okay yes

Answer 13

okay, so i made a mistake then when i was solving for 'x' quadratically becuaes i should have worked with
x2-2x+3-2=0 which, as you correctly stated, would have simplified out to x2-2x+1=0 which would have given me the factors of (x-1) (x-1) instead of (x-1) and (x+2)

Answer 14

Ok cool so we get a repeating root/solution of x=1. :) good good.

B) 1)
Hmm I think they're asking us to do a composition of functions.\[\large f(g(x))\]Sometimes its also written like this,\[\large (f \;\circ\;g)(x)\]

\[\large f(x)=x^2-2 \qquad \qquad \color{royalblue}{g(x)=2x-3}\]
Here is a quick example of how we plug things in.\[\large f(\color{orangered}{12})=(\color{orangered}{12})^2-2\]
It might be a tiny bit confusing, but we're going to plug into our function f, the entire g function.\[\large f(\color{royalblue}{g(x)})=(\color{royalblue}{2x-3})^2-2\]

Answer 15

where did the 12 come from?

Answer 16

Function notation can be confusing, was trying to give a simple example.

Answer 17

I understand the example, what you showed would mean that for example, g(x)=12, in order to plug g(x) in f, you would have to square the value of g(x) followed by a negative two. Did i interpret the example well enough?

Answer 18

The 12 was a different example, see how they're different colors? ^^ heh.

But yes, just as we can plug 12 into a function, and the 12 goes into all the x's into the function - We can also plug g(x) into the function, and g(x) will fill in all the x's within the function.

Maybe I should have written it like this first.\[\large f(\color{royalblue}{g(x)})=(\color{royalblue}{g(x)})^2-2\]Then we just replace g(x) with what we know it to be.\[\large f(\color{royalblue}{g(x)})=(\color{royalblue}{2x-3})^2-2\]

Answer 19

okay, i thought of the 12 to be different and that's why i asked where did it come from but i was not taking into consideration the colour change, lol, but i undertsand the explanation, but my concern is, what principle did you use to sqaure the value of g(x) followed by a -2?

Answer 20

okay, i have been looking carefuully at what was done and i am seeing that originally the g(x) to be plugged in f is 2x-3 and that 2x-3 would replace the 'x' in f(x), however, the 'x' in the f(x) is squared followed by a -2, so you basically just rewrote the expression by adding the square to g(x) followed by the -2 that was apart of f(x).

Answer 21

Yes good :)
As you said, the `x` in f(x) is being squared.
What we did was, we replaced the `x` with `g(x)`. So now it's g(x) being squared inside of f(x).
And then we went one step further and rewrote `g(x)` as `2x-3`.

Answer 22

Okay, for the last question i was trying to put f in g but my approach is not correct

Answer 23

Am i correct?