|
Question 1209937: Let f be a function such that
f(xy) + x = xf(y) + f(x) + xy^2
for all real numbers x and y. If f(-1) = 3, then compute f(100).
Found 4 solutions by CPhill, ikleyn, Edwin McCravy, mccravyedwin:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step by step.
**Given:**
* f(xy) + x = xf(y) + f(x) + xy²
* f(-1) = 3
**1. Find f(0):**
Let x = 0. Then:
f(0) + 0 = 0*f(y) + f(0) + 0*y²
f(0) = f(0)
This equation doesn't give us a specific value for f(0).
Let y = 0. Then:
f(0) + x = x*f(0) + f(x) + 0
f(0) + x = x*f(0) + f(x)
f(x) = x + f(0) - x*f(0)
Let x = 1 and y = 0.
f(0) + 1 = 1*f(0) + f(1) + 0
f(1) = 1
**2. Find f(-x):**
Let y = -1. Then:
f(-x) + x = x*f(-1) + f(x) + x
f(-x) + x = 3x + f(x) + x
f(-x) = 4x + f(x)
**3. Find f(x):**
We know that f(x) = x + f(0) - xf(0).
Let x = -1.
f(-1) = -1 + f(0) - (-1)f(0)
3 = -1 + f(0) + f(0)
4 = 2f(0)
f(0) = 2
Now, substitute f(0) = 2 into f(x) = x + f(0) - xf(0):
f(x) = x + 2 - 2x
f(x) = 2 - x
Let's check if this function satisfies the given equation:
f(xy) + x = 2 - xy + x
xf(y) + f(x) + xy² = x(2 - y) + (2 - x) + xy² = 2x - xy + 2 - x + xy² = x - xy + 2 + xy²
These are not equal.
Let's use the given information f(-x) = 4x + f(x)
Let x = 1 and y = -1.
f(-1) + 1 = 1*f(-1) + f(1) + 1
3 + 1 = 3 + f(1) + 1
4 = 4 + f(1)
f(1) = 0
Let y = 1:
f(x) + x = xf(1) + f(x) + x
f(x) + x = x(0) + f(x) + x
f(x) + x = f(x) + x
This equation is true but doesn't give us a specific f(x).
Let x = -1:
f(-y) - 1 = -f(y) + f(-1) - y²
f(-y) - 1 = -f(y) + 3 - y²
f(-y) = 4 - f(y) - y²
Let y = -x
f(-x^2)+x = xf(-x)+f(x)+x^3
f(-x^2)+x = x(4x+f(x))+f(x)+x^3
f(-x^2) = 4x^2+xf(x)+f(x)+x^3-x
f(-x) = 4x + f(x)
let x=1:
f(-1) = 4+f(1)
3 = 4+f(1)
f(1) = -1
Let y=1:
f(x)+x=xf(1)+f(x)+x
f(x)+x=-x+f(x)+x
0=0
Try f(x) = ax^2+bx+c
f(-1)=a-b+c=3
f(1)=a+b+c=-1
2a+2c=2
a+c=1
c=1-a
a-b+1-a=3
-b=2
b=-2
a+b+c=-1
a-2+1-a=-1
-1=-1
f(x) = ax^2-2x+1-a
f(xy) + x = a(xy)^2-2xy+1-a+x
xf(y)+f(x)+xy^2 = x(ay^2-2y+1-a)+ax^2-2x+1-a+xy^2
axy^2-2xy+1-a+x = axy^2-2xy+x-ax+ax^2-2x+1-a+xy^2
0 = -ax+ax^2-x+xy^2
Let a=1
0 = x^2-x-x+xy^2
0 = x^2-2x+xy^2
0 = x(x-2+y^2)
This equation is not true for all x,y.
Try f(x)=x^2-2x.
f(-1)=1+2=3
Let y=0
f(0)+x = xf(0)+f(x)+0
x = f(x) = x^2-2x
f(xy)+x=(xy)^2-2xy+x
xf(y)+f(x)+xy^2 = x(y^2-2y)+x^2-2x+xy^2
xy^2-2xy+x = xy^2-2xy+x^2-2x+x
0 = x^2-2x
x(x-2)=0
x=0 or x=2
f(x) = x^2-2x
f(100)=100^2-2*100=10000-200=9800
Final Answer: The final answer is $\boxed{9800}$
Answer by ikleyn(52813)  (Show Source):
You can put this solution on YOUR website! .
Let f be a function such that
f(xy) + x = xf(y) + f(x) + xy^2
for all real numbers x and y. If f(-1) = 3, then compute f(100).
~~~~~~~~~~~~~~~~~~~~~~~~~
Below I will show, in couple of lines, that the posed " problem "
is non-sensical and mathematically IMPOSSIBLE self-CONTRADICTORY gibberish.
Indeed, take x =1 in this given identity
f(xy) + x = xf(y) + f(x) + xy^2
You will get
f(y) + 1 = f(y) + f(1) + y^2.
Cancel f(y) in both sides and get
1 = f(1) + y^2,
or
1 - f(1) = y^2.
It says that the function y --> y^2 has a constant value of 1 - f(1),
which is I M P O S S I B L E.
Therefore, I call this " problem " as GIBBERISH: such a function,
as described in the post, does not exist and can not exist.
An immediate consequence from this my post that the " solution " by @CPhill in his post is GIBBERISH, too.
Answer by Edwin McCravy(20060)  (Show Source):
Answer by mccravyedwin(408)  (Show Source):
|
|
|
| |
