Question

the function f(x) = 1/(x²-3|x|+2) is discontinuous at the points​

Answer 1

The function is discontinuous at the points - 2 , - 1 , 1 , 2

Given :

$\displaystyle \sf{f(x) = \frac{1}{ ({x}^{2} - 3 |x| + 2)} }$

To find :

The points where the function is discontinuous

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf{f(x) = \frac{1}{ ({x}^{2} - 3 |x| + 2)} }$

Step 2 of 2 :

Find the points of discontinuities

$\displaystyle \sf{f(x) = \frac{1}{ ({x}^{2} - 3 |x| + 2)} }$

The function is discontinuous where the denominator of the function vanishes

So the points of discontinuities is given by

$\displaystyle \sf{ {x}^{2} - 3 |x| + 2 = 0 } \: \: \: - - - (1)$

Now two cases arise.

Case : 1 : x > 0

Then Equation 1 gives

$\displaystyle \sf{ {x}^{2} - 3 x + 2 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} - (2 + 1) x + 2 = 0}$

$\displaystyle \sf{ \implies {x}^{2} - 2x - x + 2 = 0}$

$\displaystyle \sf{ \implies x(x - 2) - 1(x - 2) = 0}$

$\displaystyle \sf{ \implies (x - 1) (x - 2) = 0}$

$\displaystyle \sf{ \implies x = 1 \: , \: 2}$

When x is positive the points of discontinuities are 1 and 2

Case : 2 : x < 0

Then Equation 1 gives

$\displaystyle \sf{ {x}^{2} + 3 x + 2 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} + (2 + 1) x + 2 = 0}$

$\displaystyle \sf{ \implies {x}^{2} +2x + x + 2 = 0}$

$\displaystyle \sf{ \implies x(x +2) +1(x + 2) = 0}$

$\displaystyle \sf{ \implies (x + 1) (x + 2) = 0}$

$\displaystyle \sf{ \implies x = - 1 \: , \: - 2}$

When x is positive the points of discontinuities are - 1 and - 2

Final answer : The function is discontinuous at the points - 2 , - 1 , 1 , 2

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