Continuity and Differentiability of piecewise functions (y=5x^2-9x, x<1; =ln(x), x>=1)

[coooool222]

This doesn't make sense, this piecewise function isnt continuous at x = 1. If it's not continuous why is the answer key telling me it is differentiable.
If a function isnt continuous it isn't differentiable.

 

[Dr.Peterson]

View attachment 34724
This doesn't make sense, this piecewise function isnt continuous at x = 1. If it's not continuous why is the answer key telling me it is differentiable.
If a function isnt continuous it isn't differentiable.

Please show us exactly what the answer key says. Perhaps you are misinterpreting it.

 

[coooool222]

Please show us exactly what the answer key says. Perhaps you are misinterpreting it.

 

[coooool222]

Wait, If a function isnt continuous in a certain point is the function as a whole still differentiable. I am confused

 

[Dr.Peterson]

View attachment 34725

Thanks. The one odd thing about this answer is that it claims the function is differentiable at x=1. Just change that from "$x\le1$"to "$x<1$" and it would be correct. Looks like a typo!

The function is differentiable everywhere except x=1, where it is discontinuous.

Wait, If a function isnt continuous in a certain point is the function as a whole still differentiable. I am confused

Of course not. But it can be differentiable everywhere else.

That's why I suggested you might be misinterpreting their answer, if you thought it claimed the function was differentiable everywhere. As it turns out, you made a correct inference from what they say, which is wrong at that one point.

 

[pka]

View attachment 34724
This doesn't make sense, this piecewise function isnt continuous at x = 1. If it's not continuous why is the answer key telling me it is differentiable.
If a function isnt continuous it isn't differentiable.

${y(x)=\begin{cases}5x^2-9x &\: x<1\\\log(x) &\:x\ge 1\end{cases}}$
Is it clear to you that $\mathop {\lim }\limits_{x \to {1^ - }} y(x) = - 4\;\& \;\mathop {\lim }\limits_{x \to {1^ + }} y(x) = 0$.
Or $y(1-)=-4\ne y(1+)=0$
Therefore, the function $y(x)$ is not continuous at $x=0$.

 

[Steven G]

The function f(x) = x^2 is differentiable everywhere.
If we define g(x) = x^2 if x is not 2, then g(x) is differentiable, except at x=2.

You are correct when you wrote if a function is not continuous, then it is not differentiable--almost

You should say if a function is not continuous at x=a, then it is not differentiable at x=a.