Question

Not able to do this right?

Answer 1

For which values of a and b is the function differentiable?

\[f(x) = ax+b for x<0\]
\[= x-x^2 for x \ge0\]

Answer 2

first off it has to be continuous to be differentiable

Answer 3

put in 0 for x and that tells you
\[b=0\]

Answer 4

sorry it has to be differentiable at x=0 idk if that changes anything

Answer 5

Yeah I can't figure out how to get a

Answer 6

yeah i got that but you have to solve for both a and b
now we know b = 0 otherwise it is not continuous

Answer 7

A cancels out when I solve for it. So does that mean it's not continous?

Answer 8

let me write this and make sure i have it correctly

Answer 9

\[f(x) = \left\{\begin{array}{rcc}
ax+b& \text{if} & x<0 \\
- x-x^2& \text{if} & x \geq 0
\end{array}
\right.
\]

Answer 10

there is no \(a\) or \(b\) in the second part?

Answer 11

no it's just \[x-x^2\]there's no negative

Answer 12

ok \[f(x) = \left\{\begin{array}{rcc}
ax+b& \text{if} & x<0 \\
-x-x^2& \text{if} & x \geq 0
\end{array}
\right.\]

Answer 13

if you plug 0 in to the first expression you get \(b\)
of you plug 0 in to the second expression you get 0
since it has to be continuous, that means \(b=0\)

Answer 14

What would a be then?

Answer 15

we didn't get there yet

Answer 16

the derivative of the first part is \(a\)
the derivative of the second part is \(1-2x\)
if you plug in 0 in to both you get \(a\) for the first one, \(1\) for the second
that tells you \(a=1\)

Answer 17

okay thanks :)