Question

For what value of the constant c is the function f continuous on (-infinity, infinity)?

Answer 1

if f(c)=(f(b)-f(a))/(b-a)

Answer 2

i mean if f'(c)

Answer 3

This is either IVT or MVT, not sure which one.

Answer 4

where c is between a and b and f(c) is between f(a) and f(b)

Answer 5

\[f(x) = cx^2 + 2x @ x<2\]

Answer 6

or f(x) = x^3-cx if x greater than or equal to two

Answer 7

f'(x) must be continuous, so:
f'(x) = 3x^2 - c, this will be continuous for any real c.

Answer 8

i was asked to find the specific value of c

Answer 9

and the answer is 2/3 but i dont know how it was arrived at

Answer 10

hmmm.. I don't remember this then. @TuringTest

Answer 11

you need to make the left and right hand limits equal at points of discontinuity

Answer 12

so look for points of discontinuity in the step function, and check the limit at each junction
in this case just x=2 is the only place to mess with

Answer 13

\[cx^2+2x=x^3-cx\] is that right?

Answer 14

at point 2, these two equations need to be equal for them to be continuous

Answer 15

\[\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=f(2)\]for it to be continuous