Question

How do you find the values of k and m such that the function is continuous?
$f\left( x \right)=\left(\begin{matrix} {{x}^{2}}+5,x > 2 \\\ m\left( x+3 \right)+k,-1 < x\le 2 \\\ 2{{x}^{3}}+x+7,x\le -1 \\\ \end{matrix}\right)$

Answer 1

Now we are given with a function. Since the function is continuous we can say that $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( -1 \right)$ and $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 2 \right)$ . Now using this condition we will form two linear equations in m and k. Then we will solve the equations simultaneously to find the value of m and k.

Complete step by step answer:
Now the function is said to be a continuous function if the graph of the function does not break. Hence simply continuous function is a function whose graph is continuous.
Now we say that a function is continuous at point a if $\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)$ .
Hence we will use this to solve the given problem and find the value of m and k.
Now consider the function.
$f\left( x \right)=\left(\begin{matrix} {{x}^{2}}+5,x > 2 \\\ m\left( x+3 \right)+k,-1 < x\le 2 \\\ 2{{x}^{3}}+x+7,x\le -1 \\\ \end{matrix}\right)$
Now since the function should be continuous we have $\displaystyle \lim_{x \to {{2}^{+}}}f\left( x \right)=f\left( 2 \right)$
Hence we can say that
$\begin{aligned} & \Rightarrow {{2}^{2}}+5=m\left( 2+3 \right)+k \\\ & \Rightarrow 9=5m+k \\\ \end{aligned}$
Hence we have $5m+k=9............\left( 1 \right)$
Now similarly we have $\displaystyle \lim_{x \to -{{1}^{-}}}f\left( x \right)=f\left( -1 \right)$
Hence using this we get,
$\begin{aligned} & \Rightarrow 2{{\left( -1 \right)}^{3}}+\left( -1 \right)+7=m\left( -1+3 \right)+k \\\ & \Rightarrow -2-1+7=-2m+k \\\ & \Rightarrow 4=-2m+k \\\ \end{aligned}$
Hence we have $-2m+k=4.........\left( 2 \right)$
Now subtracting equation (2) from equation (1) we get,
$\begin{aligned} & \Rightarrow 5m+k-\left( -2m+k \right)=9-4 \\\ & \Rightarrow 7m=5 \\\ & \Rightarrow m=\dfrac{5}{7} \\\ \end{aligned}$
Hence the value of m is $\dfrac{5}{7}$
Now substituting the value of m in equation (2) we get,

& -2\left( \dfrac{5}{7} \right)+k=4 \\\ & \Rightarrow k=4+\dfrac{10}{7} \\\ & \Rightarrow k=\dfrac{28+10}{7} \\\ & \Rightarrow k=\dfrac{38}{7} \\\ \end{aligned}$$ Hence we have $$m=\dfrac{5}{7},k=\dfrac{38}{7}$$ **Hence the function is continuous if $m=\dfrac{5}{7}$ and $k=\dfrac{38}{7}$.** **Note:** Now note that if a function is continuous then we have $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ . Here since the function is same for x = 2 and $x={{2}^{+}}$ we haven’t consider the case of $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 2 \right)$ . Similarly for - 1 we haven’t considered the case of $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( -1 \right)$ .