Continuity. Example Problem
Let f(x)={{ax^2 +3; x^3 +4x-1, x<=-2 Left; x>-2 Right For what value of a is f(x) continuous? ______________________________________________________ * lim_(x->-2)-f(x)=lim_(x->-2)+f(x)=f(-2)* lim_(x->-2)-(ax^2 +3)=4a+3=f(-2) lim_(x->-2)+(x^3 +4x-1)=-17 4a+3=-17 4a=-20 -> a=-5 This is a typical continuity question that you might see. Let f of x equal this piecewise-defined function right here. And it wants to know for what value of a will this function f of x be continuous. So we have to show that the limit as x approaches negative 2 from the left is equal to the limit as x approaches negative 2 from the right. And that has to equal the function value at negative 2. To define the limit as x approaches negative 2 from the left, I note that if x is less than or equal to negative 2, then x is left of negative 2. If x is greater than negative 2, then x is right of negative 2. This will define my left-hand limit and my right-hand limit. Returning to the example, suppose we want to evaluate the limit as x approaches negative 2 from the right-hand side. That function, then, is equal to x cubed plus 4x minus 1. Evaluating that limit leads to the conclusion that -2 cubed is equal to -8, and another -8 gives us a result of -16 minus 1, which equals -17. We have a limit from the left, a limit from the right. Looking up here, this left-hand side is attached to the function value, so we’ll make a note that says that this equals f of negative 2. Remember, it’s not enough to say that the limit from the left is equal to the limit from the right. The function must also equal that value. Now that we have shown what all these components are equal to, remember our function is continuous if these two components are the same. So 4a plus 3 is going to be set equal to negative 17. Subtracting 3 from both sides gives a equals negative 20, and so a equals negative 5.
