If f(x)=|x−1|+|x+3|, then discuss the continuity and differentiability of the function at x=-3 and x=1.

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Consider the given function  f(x)=|x−1|+|x+3|  First, check the continuity of the function at x=1  Since a function f(x) is continuous at x=a if  limx→a−f(x)=limx→a+f(x)=f(a)  Now check the continuity at the point x=1, then  limx→1−f(x)=limx→1−(|x−1|+|x+3|)  =−limx→1−(x−1)+limx→1−(|x+3|)  =0+4  =4  And  limx→1+f(x)=limx→1+(|x−1|+|x+3|)  =limx→1+(x−1)+limx→1+(|x+3|)  =0+4  =4  Therefore, the function is continuous at x=1  Now, check the continuity at the point x=-3, then  limx→−3−f(x)=limx→−3−(|x−1|+|x+3|)  =limx→−3−(|x−1|)−limx→−3−(x+3)  =(|−3−1|)−(−3+3)  =|−4|−0  =4+0  =4  And  limx→−3+f(x)=limx→−3+(|x−1|+|x+3|)  =limx→−3+|x−1|+limx→−3+|x+3|  =(|−3−1|)+(−3+3)  =4+0  =4  And

If f(x)=|x-1|+|x+3|, then discuss the continuity and differentiability of

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