Help me with this one, please?

Question

Determine the value of "c" so that f(x) is continuous on the entire real line:

f(x) = { x-2, x<=5 AND cx-3, x>5 }

I'm not really sure what it's asking.

Answer 1

For f(x) to be continuous, it must have the same value for both equations at x = 5.

If x &lt;= 5, f(x) = x-2, so f(5) = 3

If x &gt; 5, f(x) = cx - 3, and c must be chosen so that the limit of f(x) as x approaches 5 from above is 3.  (f(5) is calculated from the first equation, so you must talk about the limit as x approaches 5 from above having the same value.)

So if c*5 -3 = 3, then c = 6/5 and the second part of the function definition is

If x &gt; 5, f(x) = 6x/5 - 3, which value approaches f(5) = 3 as x approaches 5 from above.

Answer 2

Continuous= the function don&#039;t jump, then

f(y)=f(y+e) (e=very tiny number)

Solution:

(x=5) x-2=cx-3 (x=5+0.00...01)

5-2=5c-3

c=6/5

Answer 3

Start with

x-2, x&lt;=5 (since it has the equals sign included)

when x=5,

5-2=3

so then for

cx - 3  = 3 when x=5

so then

c*5 - 3 = 3

5c - 3 = 3

5c = 6

c = 6/5

Answer 4

For the solution to be continuous, there can be no break in the solution. The question is simply this: &quot;Where could f(x) break?&quot; Since each piece (x-2 and cx-3) are linear, they are both continuous throughout. Therefore, the answer is also simple: &quot;Where x-2 ends and cx-3 begins is the only place there could be discontinuity.&quot;

So, you must make sure that there is no break at the point where x=5 (the endpoint of both lines). So, x-2 must equal cx-3 at x=5 or:

5-2 = 5c-3

3 = 5c-3

6 = 5c

c = 6/5

so f(x) = { x-2, x&lt;=5 AND 6c/5-3, x&gt;5 }