Let F be a function defined as $F:[-0.5,2] \rightarrow R$ such that:
$$ f(x)=[x²-3]$$
Find the number of points where F is not continuous in $[-0.5,2]$
Let F be a function defined as $F:[-0.5,2] \rightarrow R$ such that:
$$ f(x)=[x²-3]$$
Find the number of points where F is not continuous in $[-0.5,2]$
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Write the function part wise where it breaks or becomes integral in this case.
$$f(x) = \left\{-3 ; -\frac{1}{2}<x<1 \\ -2;1<x<\sqrt2 \\ -1; \sqrt2<x<\sqrt3 \\ 0;\sqrt3<x<2 \right\}$$
So clearly the function is not continuous at $x=1,\sqrt2,\sqrt3,2$
So its 4 points.
BTW my school teacher told me this trick : $[continuous \ and \ differentiable \ function]$ is discontinuous at points where it becomes integer, but not a point of minima.
So here It was evident from graph of $y=x^2-3$ that all points where function becomes integral except $x=0$ are points of discontinuity......:-) Hope it helps....Stay Home,Stay safe