Question

Suppose f and g are continuous functions such that \(g\left( {\bf{2}} \right) = {\bf{6}}\) and \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{2}}} \left( {{\bf{3}}f\left( x \right) + f\left( x \right)g\left( x \right)} \right) = {\bf{36}}\). Fine \(f\left( {\bf{2}} \right)\).

Short answer

The value of \(f\left( 2 \right)\) is 4.

Step-by-step solution

Simplify the expression \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{2}}} \left( {{\bf{3}}f\left( x \right) + f\left( x \right)g\left( x \right)} \right)\)

As f and gare continuous everywhere, then;

\(\begin{aligned}\mathop {\lim }\limits_{x \to 2} \left( {3f\left( x \right) + f\left( x \right)g\left( x \right)} \right) &= 3f\left( 2 \right) + f\left( 2 \right)g\left( 2 \right)\\ &= f\left( 2 \right)\left( {3 + 6} \right)\\ &= 9f\left( 2 \right)\end{aligned}\)

Find the value of \(f\left( {\bf{2}} \right)\)

As the value of \(\mathop {\lim }\limits_{x \to 2} \left( {3f\left( x \right) + f\left( x \right)g\left( x \right)} \right)\) is 36, therefore;

\(\begin{array}{c}9f\left( 2 \right) = 36\\f\left( 2 \right) = 4\end{array}\)

So, the value of \(f\left( 2 \right)\) is 4.