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.
Constant number
-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$.
Function
-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 ) }$.
Product of function
Expert Answer
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 $.