16. Discuss the continuity of the function f, where f is defined by 

f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.

Answers (1)

Given function is
f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.
GIven function is defined for every real number k 
Different cases are their
case (i)  k < -1
f(k) = -2\\ \lim_{x\rightarrow k}f(x) = -2\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for every value of k < -1

case(ii)    k = -1
f(-1) = -2 \\ \lim_{x\rightarrow -1^-}f(x) = -2\\ \lim_{x\rightarrow -1^+}f(x) = 2x = 2(-1) = -2\\ R.H.L. =L.H.L. = f(-1)
Hence, given function is continous at x = -1

case(iii)  k > -1
f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for all values of  x > -1

case(vi)   -1 < k < 1
f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in -1 < k < 1 given function is continous

case(v)  k = 1
f(1) =2x = 2(1)=2\\ \lim_{x\rightarrow 1^-}f(x) = 2x=2(1)=2\\ \lim_{x\rightarrow 1^+}f(x) =2\\ R.H.L. = f(1) = L.H.L.
Hence.at  x =1 function is continous

case(vi)  k > 1
f(k) = 2 \\ \lim_{x\rightarrow k}f(x) = 2 \\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for each and every value of k > 1
case(vii)  when k < 1

f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in k < 1 given function is continuous

Therefore, continuous at all points

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