 # Suppose f and g are continuous functions such that g(2)=6 and lim[3f(x)+f(x)g(x)]=36. Find f(2), x→2

This article aims to find the value of the function $f ( x )$ at a given value. The article uses the concept of Theorem $4$. The following theorems give us an easy way to determine whether a complicated function is continuous.

-If $f ( x )$ and $g ( x )$ are continuous at $x = a$, and if $c$ is a constant, then $f ( x ) + g ( x )$, $f ( x ) − g ( x )$, $c f ( x )$, $f ( x ) g ( x )$ and $\dfrac { f ( x ) } { g ( x ) }$ (if $g ( a ) ≠ 0$) are continuous at $x = a$.

-If $f ( x )$ is continuous at $x = b$, and if $\lim {x → a g ( x ) = b }$, then $\lim {x → a f ( g ( x ) ) = f ( b ) }$.

Let:

$h ( x ) = 3 f ( x ) = f ( x ) . g ( x )$

Since $f (x )$ and $g ( x )$ are both continuous functions, according to Theorem $4$, $h ( x )$ is continuous.

$\lim _ { x \rightarrow 2 } h ( x ) = h ( 2 )$

Note that: Given that the limit in the RHS is $36$ and $g ( 2 ) = 6$.

$36 = 3 f ( 2 ) + f ( 2 ) . 6$

$36 = 9 f ( 2 )$

$f ( 2 ) = 4$

The value of the function $f ( 2 ) = 4$.

## Numerical Result

The value of the function $f (2 ) = 4$.

## Example

Suppose f and g are both continuous functions such that $g ( 3 ) = 6$ and $\lim [ 3 f ( x ) + f ( x ) g ( x) ] = 30$. Find $f ( 3 )$, $x → 3$.

Solution:

Let:

$h ( x ) = 3 f ( x ) = f ( x ) . g ( x )$

Since $f ( x )$ and $g ( x )$ are continuous, according to Theorem $4$, $h(x)$ is continuous.

$\lim _ { x \rightarrow 3 } h ( x ) = h ( 3 )$

Note: Given that the limit in the RHS is $30$ and $g ( 3 ) = 6$.

$30 = 3 f ( 3 ) + f ( 3 ) . 6$

$30 = 9 f ( 3 )$

$f ( 3 ) = 3.33$

The value of the function $f ( 3 ) =3.33$.

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