Question

Classify the discontinuities (if any) for the given function:
f(x)= { 5 x<=0
{x^2 0 14 years ago

Answer 1

at 0 and 3, both jump

Answer 2

The function is continuous for all x
b) The function has a jump discontinuity at x = 0 and 3
c) The function has a removable discontinuity at x = 3
d) The function has an infinite discontinuity at x = 3
e) The function has a removable discontinuity at x = 1

Answer 3

b is right

Answer 4

thanks Him, can you pls explain how you worked this out?

Answer 5

thanks Uzma - if you know also that owuld help too!

Answer 6

keep putting values where the function breaksa definition

Answer 7

what points are you testing?

Answer 8

the easiest way, draw the no line, mark the given piecewise function , n check the left and right side limit

Answer 9

wherever the definition of the function changes, test those pts

Answer 10

-------0---------1------------3----------------------------
5 x^2 1 x

Answer 11

how do you identify where the function changes?

Answer 12

if the value of the function is different at different sides of that point, it is discontinuous

Answer 13

it is written in the function definition...wht do they teach u at school?

Answer 14

uzma, using your line graph, how did you see the jumps?

Answer 15

excuse me Him?

Answer 16

isnt it written in ur question that there are different function for different ranges of x?

Answer 17

like for x before zero f(x) =5
for x between zero and one it is = x^2

Answer 18

so u see the function changes at x=0, from being f(x)=5 beffore it, it changes to f(x)=x^2 after it..understand?

Answer 19

yes

Answer 20

so wherever ur function changes u check for discontinuity at those points..check for values of the functn before and after tht point..if theyre unequal it is discontinuous..if the difference in the left side and right side values is finite, its a finite discontinuity, if the difference is infinite, it is the other one ie..infinite disc...

Answer 21

thank you HIM, that really made it clear! Appreciate it!

Answer 22

is that the reason why f(x) = absolute(4-x^2) has an infinite disc? at the point where x=-2?