Piecewise function continuity

[joshuamd]

I'm really not sure where to start with this.

It's worth mentioning that I'm beginning Calc 1 so I can't use L'Lopital's Rule, power rule, quotient rule, or the chain rule unfortunately.

 

[Deleted member 4993]

I'm really not sure where to start with this.
View attachment 28770
It's worth mentioning that I'm beginning Calc 1 so I can't use L'Lopital's Rule, power rule, quotient rule, or the chain rule unfortunately.

If a function f(x) is continuous at x=a,

What condition must be satisfied by the given function at x=a?​

Think about the limit of the value as x approaches 'a'.

 

[joshuamd]

Think about the limit of the value as x approaches 'a'.

It seems to me that the limit would be at 1 but I have no real reasoning for that belief. Because x can equal less than -1, -1, more than -1 but less than 1, 1, or greater than 1, there are all possible values there and that's where I am confused.

 

[Dr.Peterson]

I'm really not sure where to start with this.
View attachment 28770
It's worth mentioning that I'm beginning Calc 1 so I can't use L'Lopital's Rule, power rule, quotient rule, or the chain rule unfortunately.

This has nothing to do with derivatives, and the individual limits involved are very simple; no L'Hopital needed anywhere.

Start by finding the left- and right-hand limits at -1 and at 1. You do see that it is continuous everywhere else already, right?

 

[Deleted member 4993]

It seems to me that the limit would be at 1 but I have no real reasoning for that belief. Because x can equal less than -1, -1, more than -1 but less than 1, 1, or greater than 1, there are all possible values there and that's where I am confused.

What is:

\(\displaystyle \lim_{x \to (-1)^{-}}(6 + x)\)​

 

[joshuamd]

This has nothing to do with derivatives, and the individual limits involved are very simple; no L'Hopital needed anywhere.

Start by finding the left- and right-hand limits at -1 and at 1. You do see that it is continuous everywhere else already, right?

I know it doesn't have to do with derivates, I've just been adding that to the bottom of all of my posts.

What is:

\(\displaystyle \lim_{x \to (-1)^{-}}(6 + x)\)​

5.

 

[Dr.Peterson]

I know it doesn't have to do with derivatives, I've just been adding that to the bottom of all of my posts.

Since that was the only thinking you showed in the OP, that was all I could comment on! we'd much rather see what you do know than what you don't (or aren't allowed to use).

5.

Keep going. Find the other one-sided limits, and then think about how they apply to the problem. (Two of them will be expressions in a and b.)

And if you don't know how they apply, tell us the definitions of continuity and of limit.

 

[joshuamd]

Since that was the only thinking you showed in the OP, that was all I could comment on! we'd much rather see what you do know than what you don't

That makes sense. I'll make sure to make my future posts more informational. So I have

[math]\lim _{x\to -1^-}\left(6+x\right)=5[/math][math]\lim _{x\to -1^+}\left(ax^2+bx\right)=a-b[/math][math]\lim _{x\to 1^-}\left(ax^2+bx\right)=a+b[/math][math]\lim _{x\to 1^+}\left(ax^2+bx\right)=-1[/math]

 

[Steven G]

So what do a and b equal?

 

[joshuamd]

So what do a and b equal?

Got it. Haven't solved a system of equations in a long time so that took me a min but thank you for the help.