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  • Determine the value of ‘k’ for which the following function is continuous at x = 3: <img alt="f\left ( x \right )= \left\{\begin{matrix} \frac{\left ( x+3 \right )^{2}-36}{x-3},x\neq 3 & \\ k\, \, \, \, \, ,x= 3 & \en

Determine the value of ‘k’ for which the following function is continuous at x = 3:

f\left ( x \right )= \left\{\begin{matrix} \frac{\left ( x+3 \right )^{2}-36}{x-3},x\neq 3 & \\ k\, \, \, \, \, ,x= 3 & \end{matrix}\right.

 

 

 

 
 
 
 
 

Answers (1)

given f\left ( x \right )= \left\{\begin{matrix} \frac{\left ( x+3 \right )^{2}-36}{x-3},x\neq 3 & \\ k\, \, \, ,x= 3& \end{matrix}\right.
since f\left ( x \right ) is continuous at x = 3
\therefore \lim_{x\rightarrow 3}f\left ( x \right )= f\left ( 3 \right )
\Rightarrow \lim_{x\rightarrow 3}f\left ( x \right )=k
\Rightarrow \lim_{x\rightarrow 3}\, \frac{\left ( x+3 \right )^{2}-36}{x-3}= k
\Rightarrow \lim_{x\rightarrow 3}\, \frac{\left ( x+3 \right )^{2}-6^{2}}{\left ( x+3 \right )-6}= k \lim_{x\rightarrow 3}\, x+3+6= k
\Rightarrow 12= k thus f(x) is contineous at x =3 & k = 12

Posted by

Ravindra Pindel

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