Solving Another Functional System of Equations in Two Ways

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[damiennortier]

Replace x by g(x) in the second equation, you get f(g(x)) = 2g(x) - 7
And now, since f(g(x)) = 4x^2 - 2x + 1, we get this nice equation : 2g(x) - 7 = 4x^2 - 2x + 1
<=> 2g(x) = 4x^2 - 2x + 8
<=> g(x) = 2x^2 - x + 4

[pkmath]

Getting an inverse of f(x) would also work well. Great video Sybermath! Keep it up!

[jonsrecordcollection]

You can also exploit that f'(x) = 2, a constant. If you do the chain rule on f(g(x)) on the left hand side & the derivative on the right hand side, you get f'(g(x)) *g'(x) = 8x-2. Since f'(x) is always 2, you get 2*g'(x)=8x-2 or g'(x) = 4x-1. Integrate both sides & get g(x) = 2x^2 - x + C. Thus, 2*(2x^2-x+C)-7 = 4x^2-2x+1. Solve for C & you get C=4. Thus, g(x) = 2x^2-x+4.

[unonovezero]

You can also find f inverse (x/2+7/2) and apply it to f(g(x))

[ThAlEdison]

For this, I explicitly calculated f^-1(x)
f(x)=2x-7
f(f^-1(x))=2f^-1(x)-7
x=2f^-1(x)-7
(x+7)/2=f^-1(x)

[fivestar]

The second method is more obvious for me, because it comes from the general definition of function.

[raivogrunbaum]

Let g(x)=ax^2+bx+c then so a=2 b=-1 c=4 and g(x)= 2x^2-x+4

[titassamanta]

Let g(x)= y. So f(y)=4x^2-2x+1. That is 2y-7=4x^2-2x+1=>2y=4x^2-2x+8 => y=2x^2-x+4. Thus g(x)= 2x^2-x+4.

[eaglekingRoi]

f is a polynom with deg =1
f(g(x))= is polynom that degree is 2
So g(x) is a polynom with 2 as degree.
So we can write g(x) =2x^2 +bx+c
f(g(x)) = 2(2x^2+bx+c)-7=4x^2-2x+1
2b=-2x and 2c-7=1 so we have
b=-1 and c=4
So g(x) =2x^2-x+4

[JohnRandomness]

f(g(x)) = 2*g(x) - 7 = 4x² - 2x + 1, solve for g(x): g(x) = 2x² - x + 4

[sagnikroy]

an understandable and an easy method but long method
let g(x) = ax^2 + bx + c
fg(x) = 2(ax^2 + bx + c) - 7
fg(x) = 2ax^2 + 2bx + 2c - 7
4x^2 - 2x + 1 = (2a)x^2 + (2b)x + (2c - 7)

2a = 4, a = 2
2b = -2, b = -1
2c - 7 = 1, c = 4

therefore, g(x) = 2x^2 - x + 4

[Pizzicuddles]

Nice video. Still waiting the Sistem of Diferencial Equations :D

[rakenzarnsworld]

2(ax^2+bx+c)-7=4x^2-2x+1
a=2, b=-1, c=4
g(x)=2x^2-x+4

[nirajkumarverma]

Very interesting problem and solved in a very lucid way. I enjoyed it.👍👍👍

[lakshya]

Reading the comment section, I believe there are much more methods to solve this problem...
Either this problem is too easy or we are genius :)

[houssameoubnini]

We can do a anthor méthode is gx=f-1 0f 4x^2-2x+1 and find gx is easy

[juliocesarsilvapontes]

Solution

f(g(x)) = 4x² - 2x + 1
2g(x) - 7 = 4x² - 2x + 1
2g(x) = 4x² - 2x + 1 + 7
2g(x) = 4x² - 2x + 8
g(x) = (1/2)(4x² - 2x + 8)
g(x) = 2x² - x + 4

Verifying:
f(g(x)) = 2(2x² - x + 4) - 7
f(g(x)) = 4x² - 2x + 8 - 7
f(g(x)) = 4x² - 2x + 1 (True)

Answer
g(x) = 2x² - x + 4

[mustafasaracaloglu]

Hello. Very nice math channel. Which software and hardware do you use for your presentations ? Thank you.

[leif]

Can you post the link tonthat other similar problem? What were we tring tonsolve in that one ? F(x)?

[elias]

Plug in g into f, we get
f(g) = 2(g) - 7
But 2(g) - 7 = 4x^2 - 2x + 1
Add 7 on both sides, 2(g) = 4x^2 - 2x + 8
So g(x) = 2x^2 - x + 4
*Q*uite *e*asily *d*one!