I'm sorry, I am unable to provide a webpage title without more context.

  • Thread starter Thread starter SmittenWCalc
  • Start date Start date
  • Tags Tags
    Function
AI Thread Summary
The discussion revolves around solving a piecewise function for continuity at specific points. The user has successfully factored the first equation and determined that f(2) equals 4, leading to the equation 4 = a(4) - b(2) + 1. They simplify this to 3 = 4a - 2b but struggle to find values for a and b. Another participant suggests evaluating the function at x = 3 to create a second equation for a system of equations, which is necessary for finding the unknowns. The focus remains on ensuring the function is continuous at the transition points.
SmittenWCalc
Messages
1
Reaction score
0

Homework Statement



This is a piece wise function of course. f(x) =

(x2-4) / (x-2) if x is less than two.

ax2 - bx + 1 if x is greater than or equal to 2, or less than three.

4x - a + b if x is greater than or equal to three.

Homework Equations


The Attempt at a Solution



Alright, I know enough to factor the top of the fist equation and get x+2. That means when x is two, f(x) is four. We can use f(x) in this case because we are making the function continuous. I've gotten as far as plugging in this value in the second equation and getting

4 = a4 - b2 + 1

but I don't know what to do from here, or how to get the values of a and b. I think I subtract one from the right and get

3 = a4 - 2b

Now I am definitely stuck.
 
Physics news on Phys.org
SmittenWCalc said:

Homework Statement



This is a piece wise function of course. f(x) =

(x2-4) / (x-2) if x is less than two.

ax2 - bx + 1 if x is greater than or equal to 2, or less than three.

4x - a + b if x is greater than or equal to three.

Homework Equations


The Attempt at a Solution



Alright, I know enough to factor the top of the fist equation and get x+2. That means when x is two, f(x) is four. We can use f(x) in this case because we are making the function continuous. I've gotten as far as plugging in this value in the second equation and getting

4 = a4 - b2 + 1

but I don't know what to do from here, or how to get the values of a and b. I think I subtract one from the right and get

3 = a4 - 2b

Now I am definitely stuck.
Written in a more useful way, your equation is

4a - 2b = 3Now, what about at x = 3? You want the function to be continuous there, as well, right? What needs to happen for f to be continuous at x = 3?

That should give you another equation so that you have a system of two equations in the unknowns a and b.
 
SmittenWCalc said:

Homework Statement



This is a piece wise function of course. f(x) =

(x2-4) / (x-2) if x is less than two.

ax2 - bx + 1 if x is greater than or equal to 2, or less than three.

4x - a + b if x is greater than or equal to three.

Homework Equations


The Attempt at a Solution



Alright, I know enough to factor the top of the fist equation and get x+2. That means when x is two, f(x) is four. We can use f(x) in this case because we are making the function continuous. I've gotten as far as plugging in this value in the second equation and getting

4 = a4 - b2 + 1

but I don't know what to do from here, or how to get the values of a and b. I think I subtract one from the right and get

3 = a4 - 2b

Now I am definitely stuck.
Hello SmittenWCalc. Welcome to PF!

So, you have the following:
\displaystyle \lim_{x\to\,2-}f(x)=\lim_{x\to\,2-}(x+2)=4\ .

\displaystyle \lim_{x\to\,2+}f(x)=\lim_{x\to\,2+} (ax^2+bx+1)=4a+2b+1\ .

Do something similar at x=3 .
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top