Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Find the values of m and b that make f differentiable everywhere. f(x)=x^2, if x2. I know how to do these types of problems with only one variable, but am lost with two.

OpenStudy (freckles):

ok so we do need it to be continuous to be differentiable

OpenStudy (freckles):

so we need the left limit to equal the right limit as we approach x=2

OpenStudy (freckles):

\[\lim_{x \rightarrow 2^-}f(x)=\lim_{x \rightarrow 2^+}f(x) \\ \lim_{x \rightarrow 2^-}x^2=\lim_{x \rightarrow 2^+}(mx+b) \\ \]

OpenStudy (freckles):

evaluate the limit on both sides

OpenStudy (freckles):

and since you want it to also need the function to be smooth in order to be differentiable then you need the left derivative to equal the right derivative

OpenStudy (freckles):

\[\lim_{x \rightarrow 2^-}f'(x)=\lim_{x \rightarrow 2^+}f'(x) \\ \lim_{x \rightarrow 2^-}2x=\lim_{x \rightarrow 2^+}m\]

OpenStudy (anonymous):

aha. so we will have a system of equations

OpenStudy (freckles):

you therefore have a system of equations to solve

OpenStudy (freckles):

lol yeah

OpenStudy (anonymous):

thanks a lot

OpenStudy (freckles):

do you think you need further assistance with this one

OpenStudy (anonymous):

nah i got it thanks man

OpenStudy (freckles):

I probably need assistance in slowing down my typing so my sentences will actually make sense

OpenStudy (anonymous):

or woman lol

OpenStudy (anonymous):

nah i prefer not waiting forever

OpenStudy (anonymous):

hmmm are you sure?

OpenStudy (anonymous):

i understand them at least haha

OpenStudy (anonymous):

oh yeah, you are sure

OpenStudy (freckles):

I could be an alien without a sexual orientation

OpenStudy (freckles):

am i sure?

OpenStudy (anonymous):

a unicorn with a rainbow is most likely female

OpenStudy (anonymous):

like a truck with mudflaps

OpenStudy (freckles):

Maybe it is to deceive you

OpenStudy (anonymous):

yeah thats why i added that haha. But you never know nowadays

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!