Let's say that f and g are continuous functions, and&nbsp;g(2)=6&nbsp;and&nbsp;lim[3f(x)+f(x)g(x)]=36&nbsp;Find&nbsp;f(2),x→2

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Let&#039;s say that f and g are continuous functions, and&amp;nbsp;g(2)=6&amp;nbsp;and&amp;nbsp;lim[3f(x)+f(x)g(x)]=36&amp;nbsp;Find&amp;nbsp;f(2),x→2

crocolylec · Accepted Answer

Let&amp;nbsp;h(x)=3f(x)=f(x)⋅g(x)&amp;nbsp;Since&amp;nbsp;f(x)&amp;nbsp;and&amp;nbsp;g(x)&amp;nbsp;are continuous, by theorem 4 h(x) is continuous&amp;nbsp;That means&amp;nbsp;limx→2h(x)=h(2)&amp;nbsp;limx→23f(x)+f(x)⋅g(x)=3f(2)+f(2)⋅g(2)&amp;nbsp;Keep in mind: Given that &amp;nbsp;g(2)=6&amp;nbsp;and the maximum in the RHS is 3636=3f(2)+f(2)⋅6&amp;nbsp;36=9f(2)&amp;nbsp;4=f(2)&amp;nbsp;Result:&amp;nbsp;f(2)=4

madeleinejames20 · Accepted Answer

Answer: 4Explanation:We know that g(2)=6 and limx→2(3f(x)+f(x)g(x))=36. We&#039;re asked to find f(2).Let&#039;s start by manipulating the limit expression using algebraic operations:limx→2(3f(x)+f(x)g(x))=36We can factor out f(x) from both terms:limx→2(f(x)(3+g(x)))=36Since the limit is taken as x approaches 2, we can substitute x=2 into the expression:f(2)(3+g(2))=36Substituting the known value g(2)=6:f(2)(3+6)=36Simplifying further:9f(2)=36Now, we can solve for f(2) by dividing both sides of the equation by 9:f(2)=369=4Therefore, the value of f(2) is 4.

nick1337 · Accepted Answer

Given that f and g are continuous functions, and g(2)=6, we are given that limx→2(3f(x)+f(x)g(x))=36. We need to find the value of f(2).Let&#039;s start by manipulating the limit expression using algebraic operations. We can rewrite the expression as:limx→2(f(x)(3+g(x)))=36Since the limit is taken as x approaches 2, we can substitute x=2 into the expression:f(2)(3+g(2))=36Substituting the given value g(2)=6:f(2)(3+6)=36Simplifying further:f(2)(9)=36Dividing both sides of the equation by 9:f(2)=369=4Therefore, the value of f(2) is 4.

Don Sumner · Accepted Answer

We are given that f and g are continuous functions. Additionally, g(2) = 6. We need to find f(2) given the limit:limx→2(3f(x)+f(x)g(x))=36Since we are interested in the behavior as x approaches 2, we will rewrite the limit expression in terms of x approaching 2:limx→2(3f(x)+f(x)g(x))=limx→2f(x)(3+g(x))=36We can now use the limit definition of continuity to evaluate this expression. We know that g(2) = 6, so as x approaches 2, g(x) approaches 6. Therefore, we have:limx→2f(x)(3+g(x))=limx→2f(x)(3+6)=limx→29f(x)=36Now, we can simplify the expression:limx→29f(x)=36Dividing both sides by 9:limx→2f(x)=369=4Using the limit definition of continuity, we can conclude that f(2) = 4.Therefore, the solution to the problem is f(2) = 4.

Suppose f and g are continuous functions such that g(2)=6 and \lim[3f(x

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