Let f(x) = 2 Domain: x less than or equal to -1
= ax+b Domain: -1 less than x or less than 3
= -2 Domain: x greater or equal to 3
Find a and b such that f is continuous.
Hello, Nimmy,Originally Posted by Nimmy
I've attached a diagram to show you what you have to calculate:
In short: There are 2 points which are connected by a straight line. So you know 2 points of this line. Use the 2-point-formula of a line to get the equation.
(You should get f(x)=-x+1, -1 < x < 3)
Greetings
EB
It $\displaystyle x\not = -1, x\not= 3$ it is certainly countinous because there are constant and linear function where are always countinous.Originally Posted by Nimmy
However we need to check $\displaystyle x=-1,3$
Let us do -1 first,
By definition of countinuity we need that,
$\displaystyle \lim_{x\to -1} f(x)=f(-1)$
Thus,
$\displaystyle \lim_{x\to -1^-}f(x)=\lim_{x\to -1^+}f(x)=2$
But the limit from the left is same as,
$\displaystyle \lim_{x\to -1^-}2=2$
And the limit from the right is same as,
$\displaystyle \lim_{x\to -1^+}ax+b=-a+b$
Thus,
$\displaystyle -a+b=2$
Similarly, if we do this for $\displaystyle x=3$ we have,
$\displaystyle 3a+b=-2$
From these two equations we have,
$\displaystyle a=-1,b=1$