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

f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.

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Given function is
f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.
given function is defined for every real number k 
There are different cases for the given function
case(i)   k > 2
f(k) = k^2+1\\ \lim_{x\rightarrow k}f(x) = k^2+1\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k > 2

case(ii)   k < 2
f(k) = k^3 -3\\ \lim_{x\rightarrow k}f(x) = k^3-3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k < 2

case(iii)  x = 2

\lim_{x\rightarrow 2^-}f(x) = x^3-3 = 2^3- 3 = 8- 3 = 5\\ \lim_{x\rightarrow 2^+}f(x) = x^2+1= 2^2+1 = 4+1 = 5\\ f(2) = 2^3-3 = 8 - 3 = 5\\ f(2)=R.H.L.=L.H.L.
Hence, given function is continuous at x = 2
There, no point of discontinuity

Posted by

Gautam harsolia

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