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

f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.

Answers (1)

Given function is
f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.
GIven function is defined for every real number k 
Different cases are their
case (i)  k < 1
f(k) = 3\\ \lim_{x\rightarrow k}f(x) = 3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for every value of k < 1

case(ii)    k = 1
f(1) = 3 \\ \lim_{x\rightarrow 1^-}f(x) = 3\\ \lim_{x\rightarrow 1^+}f(x) = 4\\ R.H.L. \neq L.H.L. = f(1)
Hence, given function is discontinous at x = 1
Therefore, x = 1 is he point od discontinuity

case(iii)   1 < k < 3
f(k) = 4 \\ \lim_{x\rightarrow k}f(x) = 4\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in 1 < k < 3 given function is continous

case(iv)  k = 3
f(3) =5\\ \lim_{x\rightarrow 3^-}f(x) = 4\\ \lim_{x\rightarrow 3^+}f(x) =5\\ R.H.L. = f(3) \neq L.H.L.
Hence. x = 3 is the point of discontinuity

case(v)  k > 3
f(k) = 5 \\ \lim_{x\rightarrow k}f(x) = 5 \\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for each and every value of k > 3
case(vi)  when k < 3

f(k) = 4 \\ \lim_{x\rightarrow k}f(x) = 4\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in k < 3 given function is continous
 

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