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Problem 47 Medium Difficulty
Suppose $ f $ and $ g $ are continuous functions such that $ g(2) = 6 $ and $ \displaystyle \lim_{x \to 2} [3f(x) + f(x)g(x)] = 36 $. Find $ f(2) $.
Answer
If $f$ and $g$ are continuous and $g(2)=6,$ then $\lim _{x \rightarrow 2}[3 f(x)+f(x) g(x)]=36 \Rightarrow$
\[
3 \lim _{x \rightarrow 2} f(x)+\lim _{x \rightarrow 2} f(x) \cdot \lim _{x \rightarrow 2} g(x)=36 \Rightarrow 3 f(2)+f(2) \cdot 6=36 \Rightarrow 9 f(2)=36 \Rightarrow f(2)=4
\]
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Maitha A.
September 12, 2021
how did you get the 9 ?
Maitha A.
September 12, 2021
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Video Transcript
suppose F and G are continuous functions such that G F two is equal to six. And the limit of three times F of X plus F of X times G of X as X approaches to is 36. And here we want to find ff two. And by definition of continuity, if F and G are continuous functions, then we have the limit as X approaches to of F of X, this is equal to F F two. And the limit as X approaches to of G of X, this is equal to G F two. Now, since june of two is equal to six, then this is equal to six. And so the limit as X approaches to of three times F of X plus F of X times G fx This is equal to the limit as X approaches to of three times F of X plus, we have the limit as X approaches to of f of x times geo vex, which we can be right into three times the limit as X approaches to of F of X plus, Yes, the limit as X approaches to of F of X. This times the limit as X approaches to of G of X. Now, if we factor out the limit of F of X as X approaches to, we have limit as X approaches to of F of X this times three plus, the limit as X approaches to G fx given that this equals 36 the limits of G of X as X approaches to six, then we have 36. This is equal to the limit as X approaches to of F of X, this times three plus six. And since this is nine, we have the limit as X approaches to of F of X. This is equal to 36/9 or four. And because F of X container was then this limit of F of X as X approaches to this is equal to the value of F at two. Therefore, F of to this is equal to four.
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