Question

The function  \[f\left( x \right) = \frac{x^3 + x^2 - 16x + 20}{x - 2}\] is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as

 

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Answer 1

Here, 

\[x^3 + x^2 - 16x + 20\]
\[ = x^3 - 2 x^2 + 3 x^2 - 6x - 10x + 20\]
\[ = x^2 \left( x - 2 \right) + 3x\left( x - 2 \right) - 10\left( x - 2 \right)\]
\[ = \left( x - 2 \right)\left( x^2 + 3x - 10 \right)\]
\[ = \left( x - 2 \right)\left( x - 2 \right)\left( x + 5 \right)\]
\[ = \left( x - 2 \right)^2 \left( x + 5 \right)\]

So, the given function can be rewritten as 

\[f\left( x \right) = \frac{\left( x - 2 \right)^2 \left( x + 5 \right)}{x - 2}\]

\[\Rightarrow f\left( x \right) = \left( x - 2 \right)\left( x + 5 \right)\]

If  \[f\left( x \right)\]  is continuous at  \[x = 2\] , then

\[\lim_{x \to 2} f\left( x \right) = f\left( 2 \right)\]

\[\Rightarrow \lim_{x \to 2} \left( x - 2 \right)\left( x + 5 \right) = f\left( 2 \right)\]

\[ \Rightarrow f\left( 2 \right) = 0\]

Hence, in order to make  

\[f\left( x \right)\]  continuous at \[x = 2, f\left( 2 \right)\] should be defined as 0.