If `f(x)=|(log)_(10)x|,t h e na tx=1` `f(x)` is continuous and `f^(prime)(1^+)=(log)_(10)e` `f(x)` is continuous and `f^(prime)(1^+)=(log)_(10)e` `f(x)` is continuous and `f^(prime)(1^-)=(log)_(10)e` `f(x)` is continuous and `f^(prime)(1^-)=-(log)_(10)e`

Explanation

Checking the given function at `x=1,`

`f(x)∣_(x->1+)​=log_(10)​x=0`

and,

`f(x)∣_(x->1−)​=−log_(10)​x=0`


`f′(x)∣_(x->1+)​=d/(dx(log_(10)​x))​=1/(xlog 10)​=1/(log 10)`

So,

`f′(x)∣_(x->1−)​=d/(dx(−log_(10)​x))​=−1/(xlog 10)​=-1/(log 10)​`
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