What Values of 'a' and 'b' Ensure the Continuity of the Function f(x)?

[darshanpatel]

Homework Statement

Find the values of 'a' and 'b' that make f(x) a continuous function.

f(x) =
x+4, x≤-1
ax+bx, -1<x<3
3x+2, x≥3

Homework Equations

None

The Attempt at a Solution

I plugged -1 and 3 into their respective functions to get the points: (-1,3) and (3,11)
(-1)+4=3
3(3)+2=11

Found the slope between those two points: m=2, otherwise known as 'a'
m=(11-3)/(3-(-1))=2 ---> a

Plugged that into the point-slope formula using the point (-1,3)
y-3=2(x+1)

Solved it for b:
y-3=2x+2
y=2x+5 -----> y=mx+b so 5 is 'b'

a=2 b=5

Line that fills the gap or jump between other two equations is: y=2x+5

I even graphed it to make sure and that is the equation of the line that would fill the gap between (x+4) and (3x+2) but if you plug the values of 'a' and 'b' into the original equation, it isn't right and doesn't make sense:

ax+bx ----> 2x+5x ----> f(x)=7x to make it continues, which is wrong

I don't understand the +bx part of the ax+bx

 

[Simon Bridge]

f(x) =
x+4, x≤-1
ax+bx, -1<x<3
3x+2, x≥3

...
ax+bx ----> 2x+5x ----> f(x)=7x to make it continues, which is wrong

... shouldn't that be ax+b: -1<x<3

 

[darshanpatel]

That's what I thought it should be but it says ax+bx and I don't think it's possible with ax+bx

... shouldn't that be ax+b: -1<x<3

 

[Simon Bridge]

If you insist - then the function in the regeon we are interetsed in is:
f(x)=ax+bx=(a+b)x

But you calculated a and b for the function f(x)=ax+b ... which is a totally different function.You need f(-1)=3 and f(3)=11 so do the math for f(x)=(a+b)x and see what happens.

 

[darshanpatel]

If you insist - then the function in the regeon we are interetsed in is:
f(x)=ax+bx=(a+b)x

But you calculated a and b for the function f(x)=ax+b ... which is a totally different function.You need f(-1)=3 and f(3)=11 so do the math for f(x)=(a+b)x and see what happens.

When I try using a system of equations, it cancels all the variables out?

Such as 11=(a+b)(3)
(11/3)-b=a

That cancels out:
3=(a+b)(-1)
-3=a+b
when you plug it in

 

[Simon Bridge]

When I try using a system of equations, it cancels all the variables out?

You are asking me? It either did or it didn't - you are the one doing the problem ;)

When all the variables cancel out - what does that mean?

Note: you did not write out the calculation correctly.
You start with:
(1) a+b=-3
(2) 3(a+b)=11

using (2) gets: a=(11/3)-b
sub into (1) gets: (11/3)-b+b=-3 => 11=-9
... what does that usually mean?

Check it with a graph:
What sort of curve is f(x)=(a+b)x ?
Plot the two points you need it to pass through, then try to draw one of those curves through both of them.
What do you discover?

From your discovery, what do you conclude about the question?

 

[darshanpatel]

You are asking me? It either did or it didn't - you are the one doing the problem ;)

When all the variables cancel out - what does that mean?

Note: you did not write out the calculation correctly.
You start with:
(1) a+b=-3
(2) 3(a+b)=11

using (2) gets: a=(11/3)-b
sub into (1) gets: (11/3)-b+b=-3 => 11=-9
... what does that usually mean?

Check it with a graph:
What sort of curve is f(x)=(a+b)x ?
Plot the two points you need it to pass through, then try to draw one of those curves through both of them.
What do you discover?

From your discovery, what do you conclude about the question?

The two points it needs to go through is (-1,3) and (3,11) indicating a straight line. When the variables cancel out it indicates...? I don't recall or remember anything about learning that

 

[darshanpatel]

You are asking me? It either did or it didn't - you are the one doing the problem ;)

When all the variables cancel out - what does that mean?

Note: you did not write out the calculation correctly.
You start with:
(1) a+b=-3
(2) 3(a+b)=11

using (2) gets: a=(11/3)-b
sub into (1) gets: (11/3)-b+b=-3 => 11=-9
... what does that usually mean?

Check it with a graph:
What sort of curve is f(x)=(a+b)x ?
Plot the two points you need it to pass through, then try to draw one of those curves through both of them.
What do you discover?

From your discovery, what do you conclude about the question?

I understand that in order to solve this, you need to use a system of equations buy I am getting confused with setting it up and solving it:

x+4=y
ax+bx=y
3x+2=y

solving for 'y'
x+4=3x+2
x=1

plug 'x' in one of the equations you get: y=5

plug 'x' and 'y' into ax+bx=y

a(1)+b(1)=5
a+b=5

now I am back in the same loop but with a different approach, how do I get either 'a' or 'b'?

 

[Simon Bridge]

If you don't follow instructions or answer questions I cannot help you.

 

[darshanpatel]

You are asking me? It either did or it didn't - you are the one doing the problem ;)

When all the variables cancel out - what does that mean?

Note: you did not write out the calculation correctly.
You start with:
(1) a+b=-3
(2) 3(a+b)=11

using (2) gets: a=(11/3)-b
sub into (1) gets: (11/3)-b+b=-3 => 11=-9
... what does that usually mean?

Check it with a graph:
What sort of curve is f(x)=(a+b)x ?
Plot the two points you need it to pass through, then try to draw one of those curves through both of them.
What do you discover?

From your discovery, what do you conclude about the question?

When all the variables cancel out, I believe it is a parallel line.
f(x)=(a+b)x indicates a straight line, the two significant points have a slope of 2 between them with a y-intercept of 5. That is why I got f(x)=2x+5 at the beginning. But since it does not follow ax+bx, I tried a different approach. Which does not seem to work out very well either.

I am trying to answer all of the questions as best as I can, but I don't know some of it, sorry.

 

[Simon Bridge]

Did you figure out what sort of curve (a+b)x is? Hint: put a+b=m.
Did you do the graph thing?

 

[darshanpatel]

Did you figure out what sort of curve (a+b)x is? Hint: put a+b=m.
Did you do the graph thing?

It would be a straight line because when you plug in values for 'a' and 'b' they will add to for some 'x' such as 7x or something.

 

[Mentallic]

It would be a straight line because when you plug in values for 'a' and 'b' they will add to for some 'x' such as 7x or something.

What Simon is trying to get at is that there is something special about the particular line y=mx for some constant m. How does it differ to y=mx+b for some b that isn't 0?

 

[darshanpatel]

What Simon is trying to get at is that there is something special about the particular line y=mx for some constant m. How does it differ to y=mx+b for some b that isn't 0?

It is a y-intercept, so if it has a b, then it indicates a y intercept

 

[Mentallic]

It is a y-intercept, so if it has a b, then it indicates a y intercept

If it doesn't have a b (this is the same as b=0) then it still has a y-intercept. It's the origin or (0,0) point that all lines y=mx pass through.

You should notice some symmetry with all lines that go through the origin. Now, what is the value of f(-1)? What about f(3)? Plot those points on a graph and see for yourself whether a line that passes through the origin can also pass through those two points. Does the algebra 11=-9 tell you anything now?

 

[Simon Bridge]

- that's exactly what I'm getting at, yes.
When describing a curve, you should say everything important that you know about it ... in this case you knew it was a line and that it went through the origin ... but not the slope. So you say: "a straight line through the origin".

Get a piece of graph paper out - I don't know why you won't do this - draw in the x and y axis, get a ruler, and draw the line through the origin that also passes through both target points. The equation of line that is the solution to your problem. There is a special reason for getting you to do it graphically in this case so go do it.

I have a feeling that you would normally only get results like 11=-9 when you've made a mistake somewhere.
This time you haven't made a mistake. That is the correct result and it is telling you something very important. Probably one of the most powerful class of results in maths... so it's an important lesson. Do the graph.