Question 804453: The formula for a piecewise-defined function is given below.
Based off of the piecewise-defined function solve the equation f(x)=0
Is the function continuous on its domain? Explain why or why not. Answer by Edwin McCravy(20059) (Show Source):
The graph of that function is this:
Solve f(x) = 0
2x² = 0
x² = 0
x = 0
f(0) = 3(0)² = 0
f(x) is continuous at x = 0 because
lim f(x) = lim 3x = 0
x->0- x->0-
lim f(x) = lim 2x² = 0
x->0+ x->1+
Therefore
lim f(x) = lim f(x) = lim f(x) = f(0) = 0
x->1- x->1+ x->1
which proves that f(x) is continuous at x=0
However f(x) is not continuous at x=1 because
lim f(x) = lim 2x² = 3
x->1- x->1-
lim f(x) = lim x²-3 = -2
x->1+ x->1+
Therefore
lim f(x) ≠ lim f(x)
x->1- x->1+
Since f(x) is not continuous at x=1, it is not continuous
everywhere on its domain.
Edwin
