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

-If $ f ( x ) $ and $ g ( x )$ are**Â continuous**Â at $ x = a $, and if $ c $ is a **constant**, then $ f ( x ) + g ( x )$, $ f ( x ) âˆ’ Â g ( x )$, $ c f ( x ) $, $ f ( x Â ) g ( x )$ and $\dfrac { f ( x ) } { g ( x ) } $ (if $ g ( a ) â‰ 0$) are **continuous**Â at $ x = a$.

-If $ f ( x ) $ is **continuous**Â at $ x = b $, and if $ \lim {x â†’ a g ( x ) = b } $, then $ \lim {x â†’ a f ( g ( x ) ) = f ( b ) }$.

**Expert Answer**

**Let:**

\[ h ( x ) = 3 f ( x ) = f ( x ) . g ( x ) \]

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

\[ \lim _ { x Â \rightarrow 2 } h ( x ) Â = h ( 2 ) \]

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

\[ 36 = 3 f ( 2 ) + f ( 2 ) . 6 \]

\[ 36 = 9 f ( 2 ) \]

\[ f Â ( 2 ) = 4 \]

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

**Numerical Result**

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

**Example**

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

**Solution:**

**Let:**

\[ h ( Â x ) Â = 3 f ( x ) Â = f ( x ) . g ( x ) \]

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

\[ \lim _ { x \rightarrow 3 } h ( x ) Â = h ( 3 ) \]

Note: Given that the **limit in the RHS**Â 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**Â $ f ( 3 ) =3.33 $.