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  • Find all points of discontinuity of f, where f is defined by f x = x+3 if x<- 3 -2 x >- 3

6. Find all points of discontinuity of f, where f is defined by 

f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.

Answers (1)

best_answer

Given function is
f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.
GIven function is defined for every real number k 
Different cases are their
case (i)  k < -3
f(k) = -k + 3\\ \lim_{x\rightarrow k}f(x) = -k + 3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for every value of k < -3

case(ii)    k = -3
f(-3) = -(-3) + 3 = 6\\ \lim_{x\rightarrow -3^-}f(x) = -k + 3=-(-3)+3 = 6\\ \lim_{x\rightarrow -3^+}f(x) = -2x = -2(-3) = 6\\ R.H.L. = L.H.L. = f(-3)
Hence, given function is continuous for x = -3

case(iii)   -3 < k < 3
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 -3 < k < 3 given function is continuous

case(iv)  k = 3
f(3) = 6x+2 = 6\times3+2 =18+2=20\\ \lim_{x\rightarrow 3^-}f(x) = -2x = -2(3) = -6\\ \lim_{x\rightarrow 3^+}f(x) = 6x+2 = 6\times3+2 = 20\\ R.H.L. = f(3) \neq L.H.L.
Hence. x = 3 is the point of discontinuity

case(v)  k > 3
f(k) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each and every value of k > 3

Posted by

Gautam harsolia

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