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When is |x+1| diffrentiable
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yo all values of x except x=-1
How do you know?
it's 1
the derivative of x is 1 and the derivative of 1 is 0, add them together, you would get 1
a function is not differentiable when its graph is having a pin point
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|dw:1384837229285:dw|
More formally, a function is not differentiable at \(x=c\) if \[\lim_{x\to c^-}f'(x)\not=\lim_{x\to c^+}f'(x)\] In this case, since \(f(x)=|x+1|=\begin{cases}x+1&\text{for }x>1\\0&\text{for }x=1\\-(x+1)&\text{for }x<1\end{cases}\), you have \[f'(x)=\begin{cases}1&\text{for }x>1\\DNE&\text{for }x=1&&\text{(because the limit-condition is not met)}\\-1&\text{for }x<1\end{cases}\]
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