Find a function f(x), perhaps a piecewise function that is defined but not continuous on (-infinity, infinity) for which the function lf(x)l is both defined and continuous on (-infinity, infinity).
f(x)=
lf(x)l =
y = x^2 ?
oh, sorry, that is continuous
how about y = x for x >0
y = -x for x <0
how about something really simple, like
f(x)
= 1 for x >= 0
= -1 for x < 0
nope, also continuous
try y = x+a for x>0
and y = -x-a for x</=0
LOL, too easy Steve :)
To find a function that meets the given criteria, we need to consider a function f(x) that will ensure that the absolute value of f(x), |f(x)|, is defined and continuous.
Let's define a piecewise function as follows:
f(x) = {
-x if x < 0,
x if x ≥ 0
}
This means that for x less than zero, f(x) takes the value -x, and for x greater than or equal to zero, f(x) takes the value x.
Now, let's find the absolute value function |f(x)|:
|f(x)| = {
x if x < 0,
x if x ≥ 0
}
Since |f(x)| is equal to x for both x < 0 and x ≥ 0, it is defined and continuous on the entire domain of (-∞, ∞). This is because when transitioning from negative values to non-negative values, the function smoothly changes from -x to x without any jumps or discontinuities.
Therefore, the function f(x) = {
-x if x < 0,
x if x ≥ 0
} satisfies the criterion of having a defined and continuous absolute value function |f(x)| on the interval (-∞, ∞).