Determine the values of f(6) and g(6)

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  • Thread starter laprec
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In summary, In this problem solving, you determined that the functions $f(x)$ and $g(x)$ have a common factor of $x+1$.
  • #1
laprec
19
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Kindly help with this problem solving:

f(x) is divisible by x^2-5x-6 and g(x) is divisible by x^2-2x-3
f(x)=g(x)+2x+2
a) Determine the values of f(6) and g(6)
b) State common factor for the two functions

Attempts:
For f(x):factors are x^2-5x-6=(x-6)(x+1)

For g(x):Factors are x^2-2x-3=(x-3)(x+1)

∴for question b: the common factor will be (x+1)for the two functions

Kindly assist with question (a)
 
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  • #2
laprec said:
Kindly help with this problem solving:

f(x) is divisible by x^2-5x-6 and g(x) is divisible by x^2-2x-3
f(x)=g(x)+2x+2
a) Determine the values of f(6) and g(6)
b) State common factor for the two functions

Attempts:
For f(x):factors are x^2-5x-6=(x-6)(x+1)

For g(x):Factors are x^2-2x-3=(x-3)(x+1)

∴for question b: the common factor will be (x+1)for the two functions

Kindly assist with question (a)
What you have done so far is correct. You have shown that $f(x)$ has factors $x-6$ and $x+1$ (but it may also have other factors). From that, you can say that $f(x) = (x-6)(x+1)p(x)$, for some unknown function $p(x)$. What happens if you put $x=6$ in that equation?

You are told that $f(x) = g(x) + 2x + 2$. Once you know $f(6)$, you can put $x=6$ in that equation to get $g(6)$.

You have shown that $g(x)$ has factors $x-3$ and $x+1$ (but it may also have other factors). From that, you can say that $g(x) = (x-3)(x+1)q(x)$, for some unknown function $q(x)$. So you know that $f(x)$ and $g(x)$ have a common factor $x+1$. But could $f(x)$ and $g(x)$ have any other common factors? In other words, could those functions $p(x)$ and $q(x)$ have common factors? To rule out that possibility, you need to go back to the equation $f(x)=g(x)+2x+2.$
 
  • #3
Opalg said:
What you have done so far is correct. You have shown that $f(x)$ has factors $x-6$ and $x+1$ (but it may also have other factors). From that, you can say that $f(x) = (x-6)(x+1)p(x)$, for some unknown function $p(x)$. What happens if you put $x=6$ in that equation?

You are told that $f(x) = g(x) + 2x + 2$. Once you know $f(6)$, you can put $x=6$ in that equation to get $g(6)$.

You have shown that $g(x)$ has factors $x-3$ and $x+1$ (but it may also have other factors). From that, you can say that $g(x) = (x-3)(x+1)q(x)$, for some unknown function $q(x)$. So you know that $f(x)$ and $g(x)$ have a common factor $x+1$. But could $f(x)$ and $g(x)$ have any other common factors? In other words, could those functions $p(x)$ and $q(x)$ have common factors? To rule out that possibility, you need to go back to the equation $f(x)=g(x)+2x+2.$

Thanks a lot!
when I substituted x=6 into (x-6)(x+1)(px), f(x)=0. This make g(x)=-2x-2. I'm sure this is wrong.

- - - Updated - - -

Opalg said:
What you have done so far is correct. You have shown that $f(x)$ has factors $x-6$ and $x+1$ (but it may also have other factors). From that, you can say that $f(x) = (x-6)(x+1)p(x)$, for some unknown function $p(x)$. What happens if you put $x=6$ in that equation?

You are told that $f(x) = g(x) + 2x + 2$. Once you know $f(6)$, you can put $x=6$ in that equation to get $g(6)$.

You have shown that $g(x)$ has factors $x-3$ and $x+1$ (but it may also have other factors). From that, you can say that $g(x) = (x-3)(x+1)q(x)$, for some unknown function $q(x)$. So you know that $f(x)$ and $g(x)$ have a common factor $x+1$. But could $f(x)$ and $g(x)$ have any other common factors? In other words, could those functions $p(x)$ and $q(x)$ have common factors? To rule out that possibility, you need to go back to the equation $f(x)=g(x)+2x+2.$

laprec said:
Thanks a lot!
when I substituted x=6 into (x-6)(x+1)(px), f(x)=0. This make g(x)=-2x-2. I'm sure this is wrong.
I meant my substitution is wrong!
 
  • #4
laprec said:
when I substituted x=6 into (x-6)(x+1)(px), f(x)=0.
That should be $f(6)=0.$

laprec said:
This make g(x)=-2x-2.
No, it makes $g(6) = -2\times6 - 2.$
 
  • #5
Opalg said:
That should be $f(6)=0.$No, it makes $g(6) = -2\times6 - 2.$

Thanks a lot! I am grateful for your help. How do I determine if functions p(x) and q(x) have a common factor using f(x)=g(x)+2x+2?
 
  • #6
laprec said:
Thanks a lot! I am grateful for your help. How do I determine if functions p(x) and q(x) have a common factor using f(x)=g(x)+2x+2?

Hi laprec,

Suppose they have a common factor (x-a) for some constant a.
Then what is f(a) respectively g(a)?
Is that possible given that f(x)=g(x)+2x+2?
 
  • #7
I like Serena said:
Hi laprec,

Suppose they have a common factor (x-a) for some constant a.
Then what is f(a) respectively g(a)?
Is that possible given that f(x)=g(x)+2x+2?

Thanks I like Serena,
So since x+1 is a common factor, therefore g(-1)= (-1-3)(-1+1)q(x)=0
g(-1)=0
Therefore f(-1)= 0+2(-1)+2=0
Thus x+1 is a common factor of both functions since the remainder equal zero.
I will appreciate if you can confirm that this is order.
Thank you.
 
  • #8
laprec said:
Thanks I like Serena,
So since x+1 is a common factor, therefore g(-1)= (-1-3)(-1+1)q(x)=0
g(-1)=0
Therefore f(-1)= 0+2(-1)+2=0
Thus x+1 is a common factor of both functions since the remainder equal zero.
I will appreciate if you can confirm that this is order.
Thank you.

Yep. It is.
But then, we already knew that (x+1) was a common factor.
So we can divide left and right by (x+1) leaving p(x) = q(x) + 2, can't we?
Could there be another common factor (x-a)?
 
  • #9
I like Serena said:
Yep. It is.
But then, we already knew that (x+1) was a common factor.
So we can divide left and right by (x+1) leaving p(x) = q(x) + 2, can't we?
Could there be another common factor (x-a)?

If we divide (x-6)(x+1)p(x)=(x-3)(x+1)q(x)+2(x+1) by (x+1)

We will get (x-6)p(x)=(x-3)q(x)+2

p(x)=q(x)+2 ??

I am also struggling to see any common factor in the equation
Thanks!
 
  • #10
laprec said:
If we divide (x-6)(x+1)p(x)=(x-3)(x+1)q(x)+2(x+1) by (x+1)

We will get (x-6)p(x)=(x-3)q(x)+2

p(x)=q(x)+2 ??

I am also struggling to see any common factor in the equation
Thanks!

Nope. We can't just leave out (x-6) or (x-3).
However, suppose that p(x) and q(x) both contain a common factor (x-a).
Then what do we get if we fill in x=a in the equation?
 
  • #11
I like Serena said:
Nope. We can't just leave out (x-6) or (x-3).
However, suppose that p(x) and q(x) both contain a common factor (x-a).
Then what do we get if we fill in x=a in the equation?

(a-6)p(a)=(a-3)q(a)-2
Thanks so much but I couldn't make headway with the substitution.
 
  • #12
laprec said:
(a-6)p(a)=(a-3)q(a)-2
Thanks so much but I couldn't make headway with the substitution.

If (x-a) is a factor of p(x), then we can write p(x)=(x-a)r(x) for some unknown polynomial r(x), can't?
Now fill in x=a again...
 
  • #13
I like Serena said:
If (x-a) is a factor of p(x), then we can write p(x)=(x-a)r(x) for some unknown polynomial r(x), can't?
Now fill in x=a again...

Therefore p(a)= (x-a)r(x)=0
 
  • #14
laprec said:
Therefore p(a)= (x-a)r(x)=0

Yep! So?
 
  • #15
I like Serena said:
Yep! So?
So (x-a) is a factor!
 
  • #16
laprec said:
So (x-a) is a factor!

Not so fast!
If (x-a) is a factor of both p(x) and q(x).
Then p(a)=0 and q(a)=0, yes?
What does that mean for (a-6)p(a)=(a-3)q(a)+2?
 
  • #17
I like Serena said:
Not so fast!
If (x-a) is a factor of both p(x) and q(x).
Then p(a)=0 and q(a)=0, yes?
What does that mean for (a-6)p(a)=(a-3)q(a)+2?

Substituting p(a)=0 and q(a)=0 in the equation will give 0=2 (this cannot be true)
 
  • #18
laprec said:
Substituting p(a)=0 and q(a)=0 in the equation will give 0=2 (this cannot be true)

Indeed.
Therefore our assumption that (x-a) is a common factor of both p(x) and q(x) must be false.
This is called a proof by contradiction.
 
  • #19
I like Serena said:
Indeed.
Therefore our assumption that (x-a) is a common factor of both p(x) and q(x) must be false.
This is called a proof by contradiction.
So, we can conclude that (x+1) is the only common factor.
Wow! You are the best. Thanks for taking me through this process. I surely learned a lot
 

Related to Determine the values of f(6) and g(6)

1. What is the difference between f(6) and g(6)?

Both f(6) and g(6) are mathematical functions that involve inputting a value of 6 into the function. However, f(6) is typically used to represent a specific function, while g(6) is often used to represent a general function. In other words, f(6) is a specific output value, while g(6) could represent any output value for a given input of 6.

2. How do you determine the values of f(6) and g(6)?

The values of f(6) and g(6) can be determined by plugging the input value of 6 into the respective functions. For example, if f(x) = 2x + 3, then f(6) would be calculated as 2(6) + 3 = 15. Similarly, if g(x) = x^2, then g(6) would be calculated as 6^2 = 36.

3. Are there any restrictions or limitations to finding the values of f(6) and g(6)?

The restrictions or limitations to finding the values of f(6) and g(6) will depend on the specific functions being used. Some functions may have a limited domain (input values) or range (output values), which could affect the values of f(6) and g(6). Additionally, if the functions are complex or involve advanced mathematical concepts, it may be more challenging to determine the values of f(6) and g(6).

4. How can the values of f(6) and g(6) be used in scientific research?

The values of f(6) and g(6) can be used in scientific research to analyze data and make predictions. These values can provide insight into the behavior of certain systems or phenomena and can be used to test hypotheses and theories. For example, in physics, the values of f(6) and g(6) could represent the position and velocity of an object at a given time, which could be used to calculate its acceleration and make predictions about its future motion.

5. Can the values of f(6) and g(6) change over time?

The values of f(6) and g(6) can change over time if the functions themselves change. For example, if f(x) = 2x + 3, then f(6) will always be equal to 15. However, if the function is altered to f(x) = 3x + 2, then the value of f(6) will change to 20. Similarly, if the function g(x) = x^2 is changed to g(x) = x^3, then the value of g(6) will change from 36 to 216. Overall, the values of f(6) and g(6) are dependent on the functions being used and can change if these functions are altered.

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