Define a function by stating its rule such that f is discontinuous at x = -1, but f(-1) exists and the limit as x approaches -1 exists.
How would you solve this and problems like this?
Also, how would you solve:
Find the value of a such that f is continuous for all real numbers:
f(x) = 10-x^3 and x is less than and equal to 2
and ax^3 -1 and x > 2.
This is a piecewise function.
Thanks!
How would you solve this and problems like this?
Also, how would you solve:
Find the value of a such that f is continuous for all real numbers:
f(x) = 10-x^3 and x is less than and equal to 2
and ax^3 -1 and x > 2.
This is a piecewise function.
Thanks!
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Your first one is really easy, just define a piecewise function such that it has a hole at x = -1
EG
f(x) = { 3x + 7, if x ≠ -1
........ { 0 if x = -1
Notice 3(-1) + 7 = 4 (not 0), as long as we put any number other than 4 when x = -1
For your second question you need both parts of the function to be equal when x = 2
10 - (2)³ = a(2)³ - 1
10 - 8 = 8a - 1
3 = 8a
a = 3/8
EG
f(x) = { 3x + 7, if x ≠ -1
........ { 0 if x = -1
Notice 3(-1) + 7 = 4 (not 0), as long as we put any number other than 4 when x = -1
For your second question you need both parts of the function to be equal when x = 2
10 - (2)³ = a(2)³ - 1
10 - 8 = 8a - 1
3 = 8a
a = 3/8