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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 ) }$.

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 $.

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