2 Cal 1 problems cant figure out

  • #1
1) the given function is defined for x>0 except for x=2. Find the value to be assigned to f(2), if any, to guarantee that f is continuous at 2.

f( x ) = sin(pi x) / (x-2)

cant figure out how to change the equation so i can plug 2 in, i think im loking right past it, tried multiplying by x-2, didnt work

2) For what value of the constants a and b is the function f continuous for all x

f(x) = ( (ax-4)/(x-2) x not equal to 2
|
| b x = 2
(

i know b is equal to to, but cant figure out how to change the first equation to find a.
 

Answers and Replies

  • #2
211
0
ElectricMile said:
1) the given function is defined for x>0 except for x=2. Find the value to be assigned to f(2), if any, to guarantee that f is continuous at 2.

f( x ) = sin(pi x) / (x-2)

cant figure out how to change the equation so i can plug 2 in, i think im loking right past it, tried multiplying by x-2, didnt work

2) For what value of the constants a and b is the function f continuous for all x

f(x) = ( (ax-4)/(x-2) x not equal to 2
|
| b x = 2
(

i know b is equal to to, but cant figure out how to change the first equation to find a.
For the first question, do you mean "to guarantee that f is discontinuous at 2"? If that's what you mean, just substitute x=2 into the equation. You should be able to see right away that it cannot exist at that point. Hint: look at the denominator...
 
  • #3
221
0
For number 1 you have a removable discontinuity. You don't actually change the equation you must make f(x) into a piecewise defined function so that when x=2 the definition of f(x) is such that it is defined and so that it is continuous at x=2. So you would want to know the limit of f(x) as x approached 2. that is a 0/0 so you need to figure out what to do from here.

Good luck
 
Last edited:
  • #4
221
0
for number 2, all I see is an a in the expression you gave. I would need to know where the b is to figure that one out. Maybe you could rewrite it and make sure it is exaclty as it should be? Then I might be able to help you some more.

Regards
 
  • #5
1,444
2
can't you jsut use L'Hopital's rule for the first part
 
  • #6
221
0
stunner5000pt said:
can't you jsut use L'Hopital's rule for the first part
Yeah but he was suppose to figure that one out on his own...... :wink:
 

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