Asked by Anonymous
Consider the function f(x)=
x^2+a if x<0,
2x+a^2 if 0 less than or equal to x less than or equal to 1,
3 if x>1
a. determine the value(s) of a that will make f(x) continuous at x=0
b. determine the value or values of a that will make f(x) continuous at x=1
x^2+a if x<0,
2x+a^2 if 0 less than or equal to x less than or equal to 1,
3 if x>1
a. determine the value(s) of a that will make f(x) continuous at x=0
b. determine the value or values of a that will make f(x) continuous at x=1
Answers
Answered by
bobpursley
I will be happy to critique your thinking.
Answered by
Anonymous
For part a I took the limit as x tends to zero from the right of 2x+a^2. I put zero in for x and then said a=0 which would also make the limit as x approaches zero from the left of x^2+a 0 so the function would be continuous.
I don't know how to start part b because there is no variable for x>1
I don't know how to start part b because there is no variable for x>1
Answered by
bobpursley
You are correct, part of 3 is missing.
Answered by
Anonymous
Maybe the limit of x approaches 1 from the right of 3 = 3. And then set the limit of x approaches 1 from the left of 2x+a^2 = 3 and then solve for a and get 1.
Answered by
bobpursley
I am not certain I would make that assumption on the problem. However, you could be right.
Answered by
Anonymous
one more question.
if possible determine the value(s) of a that will make f(x) continuous on (negative infinity, infinity)
if possible determine the value(s) of a that will make f(x) continuous on (negative infinity, infinity)
Answered by
MathMate
There is more than one value of a that will make the function continuous at x=0, since at x=0,
x^2+a =2x+a^2
or setting x=0,
a-a²=0
a(1-a)=0
giving
a=0 or a=1.
Now you can continue with the last part.
x^2+a =2x+a^2
or setting x=0,
a-a²=0
a(1-a)=0
giving
a=0 or a=1.
Now you can continue with the last part.
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