Limits and continuity: f(2)+3=f(0), f continuous at x=1

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jmota

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Im not sure how to solve this problem, not even where to begin.

Find constants a and b such that f(2)+3=f(0) and f is continuous at x=1.

f(x)={ ax+b if x>1
3 if x=1
x^(2)-4x+b+3 if x<1
 
Re: Limits and continuity

jmota said:
Im not sure how to solve this problem, not even where to begin.


Find constants a and b such that f(2)+3=f(0) and f is continuous at x=1.

f(x)={ ax+b if x>1
3 if x=1
x^(2)-4x+b+3 if x<1
Click to expand...

What are the conditions of continuity of a function at a given point.
 
Re: Limits and continuity

* f(c) is defined
* lim f(x) exists
(x to c)
*lim f(x)=f(c)
(xto c)
 
Re: Limits and continuity

We are given that
\(\displaystyle f(x) = \left\{ {\begin{array}{rr} {ax + b,} & {x > 1} \\ {3,} & {x = 1} \\ {x^2 - 4x + b + 3,} & {x < 1} \\ \end{array} } \right.\).

Therefore: \(\displaystyle \lim _{x \to 1^ + } f(x) = a + b\;\& \,\lim _{x \to 1^ - } f(x) = b\).

Now you put it all together.
 
Re: Limits and continuity

whole reason i asked is because i dont understand the concept of the questions so this answer doesnt help me at all. So by saying "put them all together", ill just ask "put what together". so yes im still lost
 
Im trying to solve this problem but Im getting no where. I dont understand the concept of the problem at all. the question is:

Find constants a and b such that f(2)+3=f(0) and f is continuous at x=1

f(x)={ ax+b if x>1
3 if x=1
x^(2)-4x+b+3 if x<1

How do I solve the equation?
 
Re: Limits and continuity

jmota said:
whole reason i asked is because i dont understand the concept of the questions so this answer doesnt help me at all. So by saying "put them all together", ill just ask "put what together". so yes im still lost
Click to expand...
Then it is time for you to have a sit-down with your instructor.
It hopeless for you to think that the sort of confusion can be address online.
 
Re: Limits and continuity

wait. do I solve it by setting up the equations equal to 3 where x=1. for instance would i make (x^(2)-4x+b+3) into (1^(2)-4(1)+b+3=3. Which would make b=3. Then by making ax+b into a(1)+3=3, a would equal 0.
 
Re: Limits and continuity

I think I solved it a=o,b=3. I did this by setting f(1)=x^(2)-4x+b+3=3. This makes the problem (1^(2)-4(1)+b+3=3), giving 1-4+3+b=3, which leaves b=3. Then by doing the same to ax+b, I get a(1)+3=3, this makes a=0.
 
Re: Limits and continuity

jmota said:
wait. do I solve it by setting up the equations equal to 3 where x=1. for instance would i make (x^(2)-4x+b+3) into (1^(2)-4(1)+b+3=3. Which would make b=3. Then by making ax+b into a(1)+3=3, a would equal 0.
Click to expand...

You had put it together correctly - now doesn't it feel better that you figured it by yourself!!!
 
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