Help with limits?
Find a and b
f(x)= {2, x<1
{ax+b, -1
the less than sign for x<1 is actually a less than or equal to and the greater than sign for x>3 is actually a greater than or equal to
in the back of my book the answer says a=-1 and b=1 though..
Is x<1 or -1?
is the function, though piecewise, supposed to be continuous?
x is less than or equal to -1 and x is greater than or equal to 3
oh ok that's why it didnt work
I was using 1
hmmm \(\large f(x) = \begin{cases} 2, &x \le -1\\ ax+b, & -1 < x < 3\\ -2, & x \ge 3 \end{cases}\) ?
so 2=a(-1)+b 2=-a+b 2+a=b so -2=a(3)+(2+a) -4= 3a+a -4=4a a=-1 Now it works
2+-1= 1=b
Now does it make sense?
hmmm, I am assuming all that's to make it continuous... hmm
Yes it's a continuous extension type of thing
I don't understand how you knew what to do
like how did you know to set 2 to ax+b?
Well, the point is to make the function continuous at X=-1 and 3. So you want to find the values of a and b that will complete the function and "unite" the 3 smaller functions.
wait so the -1 and the -3 are pretty much what x is approaching?
Exactly
|dw:1378412017785:dw| I assume is \(f(x) = \begin{cases} 2, &x \le -1\\ ax+b, & -1 < x < 3\\ -2, & x \ge 3 \end{cases}\) thus, ax+b is really the equation a line, a line that goes from that point ON THE UPPER-LEFT to that other point ON THE LOWER-RIGHT
anyhow, once you have 2 points for it, you can pretty much get the equation for the line, using the point-slope equation
you don't need to use the point-slope equation.. but thanks
The point of this is to be able to solve the problem algebraically @jdoe0001, without the use of graphs.
right, I saw that :)
Awks I didn't see you were still on the problem - just ignore the message I sent. How did you get b?
@jdoe0001 @TheDoctorRules answer her, please
@dan815
I'm with jdoe0002 (XD) . I'm stick with slope of the line. However, your problem need the condition of "f(x) is continuous at the interval (-1,3) from then you can find a =-1 , b =1
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