Question

Use calculus to show that f(x) is continuous and differentiable at x = 1. f(x) is a piecewise function.

f(x) = { -x^3 + x for x<1 , -2x + 2 for x>= 1 ,

Answer 1

find the derivative for each piece of the piecewise function at x=1. If they are the same then the overall function is continuous and differentiable.

Answer 2

at x=1, f(1) = 0 for each piece
at x=1, f'(1) = -2 for each piece

Answer 3

f'(x) = -3x^2 + 1 for x <1
-2 for x > = 1

Answer 4

yes but how do you show it is differnetiable at x = 1?

Answer 5

there are two different branches

Answer 6

it is sufficient to show it is differentiable at x =1, that implies continuity

Answer 7

dumbcow, can you make that argument more rigorous

Answer 8

you just need to show that the derivative at x=1 is the same for both pieces and that the value of the function is the same for both pieces at x=1. plug in x=1 into your derivatives that you have performed -3*(1)^2 +1 = 2

Answer 9

equals -2
oops

Answer 10

@perl I'm giving you a medal because I love your Red Spy TF2 picture okay bro.

Answer 11

thanks

Answer 12

what do you mean for both pieces?

Answer 13

Yea dawg but Blu Spy is better

Answer 14

to show continuity...the limit as x->1 from either side should approach the same value
to show differentiability....the limit of the slope of f(x) as x->1 from either side should approach the same value

Answer 15

so f ' (x) is the limit ?

Answer 16

the piecewise function is comprised of two functions (i was using the word pieces for that)

Answer 17

then we have to show lim h->0 f(a+h) - f(a) / h

Answer 18

yes, im just confused about the endpoints