Question

How do I find a and b so that the function is differentiable?

Answer 1

what is the function?

Answer 2

It has to be differentiable at \[\Pi/4\]
f(x)= \[\cos x: for, x \le \pi/4\]
\[=a+bx :for, x> \pi/4\]

Answer 3

f is diff. when right lim and left lim exist and are equal
Hence, at \(\pi/4\), left limit is
\(lim_{x\rightarrow (\pi/4)^-} cos x =\sqrt 2/2\)

Answer 4

right limit is \(\lim_{x\rightarrow (\pi/4)^+} a+bx = a+b\sqrt 2/2\)

Answer 5

let them equal, you have \(a+b\sqrt 2/2=\sqrt 2/2\)
hence if \(a= (\sqrt 2/2)(1-b)\), then right lim = left lim , hence the function is continuous, then diffrentiable

Answer 6

wait, why didn't you plug in pi/4 for x in (a+bx)?

Answer 7

I did :)

Answer 8

oh, mistake

Answer 9

but isn't that cos (pi/4)?

Answer 10

it is \(a+ b \pi/4 = sqrt2 /2\) I am sorry

Answer 11

its ok

Answer 12

but the method is the same, just consider when a = something *b to have the function continuous,

Answer 13

next is: if it is conti.
then take derivative and let them equal

Answer 14

so sqr (2)/2= a +b (pi/4)

Answer 15

-sinx = b , when x = pi/4 , then b = -sqrt2 /2

Answer 16

then plug back to find a

Answer 17

the limit steps is for making sure that the function continuous only.

Answer 18

but you need that step to find derivative of f.

Answer 19

\[y= a + (\frac{ -\sqrt{2} }{ 2 })\left( \frac{ \pi }{ 4 } \right)= \frac{ \sqrt{2} }{ 2 }\]
is what I'm getting. Solving for a will give me my answer for b, am I correct?