If |f| is continuous at c, does it follow that f is continuous at c? Justify your answer.?

Question

Let f: D-->R and define |f|:D-->R by |f|(x)=|f(x)|. Suppose that f is continuous at c belonging to D. Prove that |f| is continuous at c. If |f| is continuous at c, does it follow that f is continuous at c? Justify your answer.

Answer 1

First, suppose f is cotinuous at c. According to the definition of continuity, for every eps >0 there exists a neighborhood U of C such that

|f(x) - f(c)| < eps (1)

for every x in U inter D. According to the triangle inequality, for such elements x we also have

|f(x) - f(c)| >= | |f(x) - |f(c) | (2) Combining (1) and (2), if follows that

| |f(x) - |f(c)| | < eps, valid for every x in U inted D, which is exactly th condition of continuity of |f| at c.

But the converse is not true. Take D = R, c = 0 and let f be given by

f(x) = -1 if x <0
f(x) = 1 if x >=0. Then, f is discontinuous at x = 0, but since |f(x)| =1 for every real x, if follows |f| is continuous at x = 0.

But if |f| is continuous at c and f(c) = 0, then f is continuous at c, as we can readily see.

Answer 2

Consider f(x)=-1 if x<0 and f(x)=1 if x>=0.
|f|(x)=1 is continuous yet f is not continuous at x=0.

Answer 3

situation A isn't there. i assume you in simple terms want solutions for B and C. In those 2, there are not any countless discontinuities (divide by utilising zeroes), and f(x) is defined for all x (i.e., area is all genuine numbers). the sole element left is to look on the barriers the place the applications substitute. B. For x=one million, f(x) = one million + 2 = 3. For x = one million+ (very incredibly extra desirable than one million), f(x) = 4 - (one million+) = 3-. interior the cut back, as x techniques one million from the splendid, f(x) techniques 3. The function is non-supply up at x=one million. C. If x = -2, f(x) = (-2)^2 = 4. For x = -2- (very incredibly under -2), f(x) = x = -2- ? 4. it somewhat is a step discontinuity at x = -2.

Answer 4

No. Take the simple case
f(x) = 6 if x>=0 and f(x)=-6 if x<0
f(x) is not continuos at 0 while If(x)I =6 is continuos