The number of points of discontinuity of the function f(x)=[x^22]-[x],x [0,4], where [ ] d…

Question

The number of points of discontinuity of the function $f(x)=[\frac{{x}^{2}}{2}]-[\sqrt{x}],x\in [0,4]$, where $[\cdot ]$ denotes the greatest integer function, is \_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

To determine the points of discontinuity of the function $f(x)=[\frac{{x}^{2}}{2}]-[\sqrt{x}]$, where $[\cdot ]$ denotes the greatest integer function, we need to identify possible values of $x$ where discontinuities might occur within the interval $[0,4]$.

Discontinuity Analysis

For the term $[\frac{{x}^{2}}{2}]$:

The probable values of $x$ that could cause discontinuities are the roots or specific values where the integer part changes between consecutive integers. The transitions happen when:

$\begin{gathered} =1,2,3,4,5,6,7,8 \ \, \implies \, x=\sqrt{2},2,\sqrt{6},2\sqrt{2},\sqrt{10},2\sqrt{3},\sqrt{14},4 \end{gathered}$

For the term $[\sqrt{x}]$:

The values of $x$ where $[\sqrt{x}]$ changes are straightforward. They occur at:

$x=1,2$

Discontinuity Check
By evaluating $f(x)$ at all these potential points, we find the function is indeed discontinuous at:

$x=1,\sqrt{2},2,\sqrt{6},2\sqrt{2},\sqrt{10},2\sqrt{3},\sqrt{14}$

Thus, the function $f(x)$ has 8 discontinuities on the interval $[0,4]$.

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