Find a value of c that makes the function continuous at the given value of x. If it is impossible, state this.
-$f ( x ) = \left\{ \begin{array} { c c }
\frac { x + 3 } { 5 | x + 3 | ^ { \prime } } & x \neq - 3 \
c , & x = - 3
\end{array} ; x = - 3 \right.$
A) 5
B) Impossible
C) -3
D) 3

Question

Find a value of c that makes the function continuous at the given value of x. If it is impossible, state this.
-$f ( x ) = \left\{ \begin{array} { c c } 
\frac { x + 3 } { 5 | x + 3 | ^ { \prime } } & x \neq - 3 \
c , & x = - 3
\end{array} ; x = - 3 \right.$
A) 5
B) Impossible
C) -3
D) 3

Quizplus · Accepted Answer

For the function to be continuous at $x = -3$, the limit as $x$ approaches $-3$ must equal the value of the function at $x = -3$, which is $c$. However, the expression $5 | x + 3 | ^ { \prime }$ involves the derivative of the absolute value function, which is not defined at $x = -3$ because the absolute value function has a corner at $x = -3$, making its derivative undefined at that point. Therefore, it is impossible to find a value of $c$ that makes the function continuous at $x = -3$.

Find a Value of C That Makes the Function Continuous f(x)={x+35∣x+3∣′x≠−3c,x=−3;x=−3f ( x ) = \left\{ \begin{array} { c c } \frac { x + 3 } { 5 | x + 3 | ^ { \prime } } & x \neq - 3 \\c , & x = - 3\end{array} ; x = - 3 \right.f(x)={5∣x+3∣′x+3​c,​x=−3x=−3​;x=−3

Find a Value of C That Makes the Function Continuous $f ( x ) = \left\{ \begin{array} { c c } \frac { x + 3 } { 5 | x + 3 | ^ { \prime } } & x \neq - 3 \\c , & x = - 3\end{array} ; x = - 3 \right.$