Question

Given the piecewise function...

Answer 1

yes @blahblahbro

Answer 2

f(x) = { 2x+a when x<=1
bx^2 - 1 when x > 1 }

For what values of a and b is f(x) differentiable at x = 1?

Answer 3

first hint :-
the function need to be continues

Answer 4

It doesn't have to be continuous

Answer 5

see check limit from right and left at x=1

Answer 6

if the function is differentiable , then its continues , thats why it's need to be cont

Answer 7

I tried doing

lim(2x+a) = lim(bx^2 - 1)
as x approaches 3

I got a = b-3

but that doesn't help

Answer 8

well ur mistack is as x approaches 3 , it should be 1

Answer 9

Oh i mistyped it. I actually took the limit as it approaches 1
I still got a = b-3

Answer 10

haha

Answer 11

thats first equation

Answer 12

nw other hint
f'(1) from right is the same from left

Answer 13

So d/dx(2x+a) = d/dx(bx^2 -1)
?

Answer 14

yes at x=1

Answer 15

So that means 2 = 2b -1
b = 3/2
?

Answer 16

Because b=3/2 is not any of the answer choices...

Answer 17

d/dx(2x+a) = d/dx(bx^2 -1) at x=1

2=2bx
b=1 not 3/2

Answer 18

not that d/dx(-1)=0 not -1

Answer 19

Oh wow I can't believe I messed up that >.<
b = 1 is an answer choice lol

Answer 20

b=1 and a=-2

Answer 21

:)

Answer 22

Thanks for your help!

Answer 23

np :)

Answer 24

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Answer 25

@blahblahbro

Answer 26

why r u so rude? @blahblahbro

Answer 27

because