I need clarity

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khris le

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1649956577537.png
This picture is a composite function question. The solution for it is in red.
My answer would be (g(f(x))) = (4x+3)^3
= 16x +9

I don't understand why the 16 x is cubed or there is a 24x
 
khris le said:
View attachment 32133
This picture is a composite function question. The solution for it is in red.
My answer would be (g(f(x))) = (4x+3)^3
= 16x +9

I don't understand why the 16 x is cubed or there is a 24x
Click to expand...
Well [imath](4x+3)^2=[(4)^2]x^2+(2)(4)(3)x+(3)^2=16x^2+24x+9[/imath]
[imath]g\circ f(x)=g(f(x))=(4x+3)^2[/imath]
 
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pka said:
Well [imath](4x+3)^2=[(4)^2]x^2+(2)(4)(3)x+(3)^2=16x^2+24x+9[/imath]
[imath]g\circ f(x)=g(f(x))=(4x+3)^2[/imath]
Click to expand...
sorry I still don't understand. So I get that I had to square the 4 and x separately now but where did the + (2) (4) (3)x come from?
 
[imath](a+b)^=a^2+2ab+b^2[/imath] ,
If [imath]g(x)=3x^2-4x+5)[/imath] then [imath]g(7)[=3(7^2)-4(7)+5=3(49)-28+5[/imath] ,also [imath]\color{blue}g(f)=3f^2-4f+5[/imath]
Then if [imath]f(x)=3x-2[/imath] then [imath]g\circ f(x)=\color{blue}g(f(x))=3(3x-2)^2-4(3x-2)+5[/imath]
 
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khris le said:
sorry I still don't understand. So I get that I had to square the 4 and x separately now but where did the + (2) (4) (3)x come from?
Click to expand...
[math](4x+3)^2=(4x+3)(4x+3)\stackrel{FOIL}{=}16x^2+12x+12x+9=16x^2+24x+9[/math]
 
pka said:
FOIL: a very thin sheet of metal, especially used to wrap food in to keep it fresh:
Click to expand...
A foil is one of the three weapons used in the sport of fencing, all of which are metal. It is flexible, rectangular in cross section, and weighs under a pound. As with the épée, points are only scored by contact with the tip, which, in electrically scored tournaments, is capped with a spring-loaded button to signal a touch. A foil fencer's uniform features the lamé (a vest, electrically wired to record hits). The foil is the most commonly used weapon in competition
 
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FYI: here's a thread about whether or not FOIL is good

I have never used FOIL myself. I don't have a strong opinion about its use. Hopefully this post might make things clear if anyone is confused about some of the posts above :).
 
FOIL has nothing to do with math, at least it shouldn't.
Students have enough formulas to know and this one is not a good one to know.
 
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Oops! ? Posted this in the wrong thread! ?

I have never taught FOIL (& never will); it's far too restrictive! I have always taught: Multiply every term in the second bracket by each term in the first bracket in turn; then gather like terms. Doesn't matter how may terms are in either bracket then!
(Just my tuppenceworth. ?)
 
khris le said:
Thank you, so I can use the method you wrote or the (a+b)^2 = a^2 = 2ab-b^2 rule?
Click to expand...
Please review your post (#6) for typos.

The correct representation of the rule - as shown in post #11 - is:

(a + b)^2 = a^2 + 2ab + b^2

Read it carefully. I missed your typos at the first read.
 
khris le said:
so I can use the method you wrote [or another method I've seen]
Click to expand...
In math, you're free to choose any method that works for you, khris (unless an exercise specifically instructs otherwise).

In the beginning, use methods that you can understand/remember, and then allow your work to evolve over time as you gain new insights and perspectives.

?

[imath]\;[/imath]
 
khris le said:
View attachment 32133
This picture is a composite function question. The solution for it is in red.
My answer would be (g(f(x))) = (4x+3)^3
= 16x +9

I don't understand why the 16 x is cubed or there is a 24x
Click to expand...
Please read my post (#14) above first.
(This method will work no matter how many terms are within the brackets or, indeed how many brackets there are!)

You understand that \(\displaystyle (g of f(x))=(4x+3)^2\) (not \(\displaystyle (4x+3)^3\) as you typed) Yes?

Then:-
\(\displaystyle (4x+3)^2=(4x+3)(4x+3)\)​

Multiplying everything in the second bracket by the first term in the first bracket (ie: \(\displaystyle 4x\)) we get:-

\(\displaystyle 4x(4x+3)=16x^2+12x\)​

Then multiplying everything in the second bracket by the second term in the first bracket (ie: 3) we get:-

\(\displaystyle 3(4x^2+3)=12x+9\)​

The final result is found when those two results are added together, so:-

\(\displaystyle (4x+3)^2=16x^2+12x+12x+9\)​

Now gathering all the like terms together (ie: all the \(\displaystyle x^2\) terms, all the \(\displaystyle x\) terms and all the constants) you end up with:-

\(\displaystyle 16x^2+24x+9\)​

But you wouldn't normally split it into two separate lines (as I have done above to illustrate each stage). You would just write:-

\(\displaystyle (4x+3)^2=(4x+3)(4x+3)=16x^2+12x+12x+9=16x^2+24x+9\)​

Does that make things any clearer for you? ?

You can extend this method (as outlined at Post #14) to deal with brackets containing more than two terms and also to deal with more more than two brackets (just deal with them two at a time, add together all the results and gather all the like terms), eg:-

\(\displaystyle (4x+3)(2x^2+4x+3)=8x^3+16x^2+12x\) plus \(\displaystyle 6x^2+12x+9=8x^3+22x^2+24x+9\)​
                                    └───────┬───────┘     └─────┬─────┘
                        Multiplying by \(\displaystyle 4x\) and Multiplying by 3
 
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