for what value of the constant C is the function F continuous on (-infinity,+infinity) where...

f(t)= {ct+8 if t (-infinity, 9] }

and f(t)= {ct^2 if t (9, infinity) }

i have no clue as to how im supposed to approch this problem. anyhelp is appreciated. thanks

Update:

CRAP for the second condition i meant ct^2-8, not ct^2

Relevance

In English, a continuous function has no gaps, so you want

ct + 8 = c*t^2 at t = 9, or

9*c + 8 = c*81

72*c = 8

c = 1/9

• Anonymous

To make the function continuous, you have to make sure that the values are the same on either side of the break point. The break point in this case is, obviously, 9. Call the left side of the function f-(t) and the right side f+(t), then:

f-(9) = c(9^2) = 81c

f+(t) = c(9) + 8 = 9c + 8

Set them equal:

81c = 9c + 8

72c = 8

c = 1/9

So there's your value. Note, please, that this only makes the function continuous. It is probably not differentiable at that point.

• locuaz
Lv 7

of ct+8=ct^2-8

ct^2-ct-16=0

and t=9

so c81 -c9-16=0

c(81-9)=16

c=16/72=8(2)/9(8)=2/9

Source(s): h