Continuity

[steph7]

Solve for the constants a and b that make the piecewise function continuous for all real numbers. f(x)={ 4-2x-x^2, x<-1 ax+b, -1 ≤x ≤1 x^3, x>1

 

[Dr.Peterson]

There are three "pieces" here: x<-1, -1≤x≤1, and x>1. They meet at -1 and at 1. What you need to do is to adjust the values of a and b so that the function will be continuous there.

What have you tried, and where are you stuck? We'd like to help you at your point of need, not just do it for you.

 

[steph7]

I understand that part. Normally when I do questions like this I’d find what f(-1) is and then find the limits x->1- and x->1+ but I can’t do that with this one because of ax+b, -1 ≤x ≤1. So how do I start the problem?

 

[pka]

Solve for the constants a and b that make the piecewise function continuous for all real numbers. f(x)={\\ 4-2x-x^2, x<-1 ax+b, -1 ≤x ≤1 x^3, x>1

\(f(x)=\begin{cases}\\4-2x-x^2 &: x<-1 \\ ax+b &: -1\le x\le 1\\ x^3 &: 1< x\end{cases}\)
\(\mathop {\lim }\limits_{x \to {-1^ - }} f = 5\;\& \;\mathop {\lim }\limits_{x \to {-1^ + }} f = - a + b\)
\(\mathop {\lim }\limits_{x \to {1^ - }} f = a+b\;\& \;\mathop {\lim }\limits_{x \to {1^ + }} f = 1\)

 

[HallsofIvy]

Do you know the distinction between taking the limit "from below" and "from above"?

Since \(\displaystyle f(x)= 4- 2x- x^2\) for x less than -1, the limit "as x approaches -1 from below" is \(\displaystyle \lim_{x\to-1} 4- 2x- x^2\) which, since \(\displaystyle 4- 2x- x^2\) is a continuous polynomial, is \(\displaystyle 4- 2(-1)- (-1)^2= 4+ 2+ 1= 7\).
Since \(\displaystyle f(x)= ax+ b\) for x between -1 and 1, the limit "as x approaches -1 from above" is \(\displaystyle \lim_{x\to-1} ax+ b\) which, since \(\displaystyle ax+ b\) is a continuous polynomial, is \(\displaystyle -a+ b\). So we have -a+ b= 7.

Do the same with the limits "from below" and "from above" at x= 1 to get a second equation in a and b and solve the two equations for a and b.

 

[pka]

Since \(\displaystyle f(x)= 4- 2x- x^2\) for x less than -1, the limit "as x approaches -1 from below" is \(\displaystyle \lim_{x\to-1} 4- 2x- x^2\) which, since \(\displaystyle 4- 2x- x^2\) is a continuous polynomial, is \(\displaystyle \huge{4- 2(-1)- (-1)^2= 4+ 2\color{red}+ 1= 7}\).

Sign error?

 

[Dr.Peterson]

I understand that part. Normally when I do questions like this I’d find what f(-1) is and then find the limits x->1- and x->1+ but I can’t do that with this one because of ax+b, -1 ≤x ≤1. So how do I start the problem?

You didn't indicate why that stopped you; I'm not sure what it is.

You can still find the limit as x approached -1 from above; since ax + b is continuous, that is just a(-1) + b = -a + b. Set that equal to the limit from above. Do the same at +1, and you'll have two equations in a and b ...