consider the function ( f(x)=\frac{2}{2x - 1} ). is there a value of ( 0lt clt3 ) such that (…

consider the function ( f(x)=\frac{2}{2x - 1} ). is there a value of ( 0lt clt3 ) such that ( \frac{f(3)-f(0)}{3 - 0}=f^{prime}(c) )? if there is, enter your answer below (write dne if there is no value of ( c )).\nquestion 3\nlet ( f(x) ) be a continuous and differentiable function for all ( -inftylt xltinfty ). suppose ( f(1)=5 ) and ( f^{prime}(x)geq3 ) for all ( -inftylt xltinfty ).\n1. what is the smallest possible value for ( f(7) )?\nanswer:\n2. what is the largest possible value for ( f(-2) )?\nanswer:

consider the function ( f(x)=\frac{2}{2x - 1} ). is there a value of ( 0lt clt3 ) such that ( \frac{f(3)-f(0)}{3 - 0}=f^{prime}(c) )? if there is, enter your answer below (write dne if there is no value of ( c )).\nquestion 3\nlet ( f(x) ) be a continuous and differentiable function for all ( -inftylt xltinfty ). suppose ( f(1)=5 ) and ( f^{prime}(x)geq3 ) for all ( -inftylt xltinfty ).\n1. what is the smallest possible value for ( f(7) )?\nanswer:\n2. what is the largest possible value for ( f(-2) )?\nanswer:

Answer

Explanation:

Step1: Apply the Mean Value Theorem (MVT)

The Mean Value Theorem states that if (y = f(x)) is continuous on the closed interval ([a,b]) and differentiable on the open interval ((a,b)), then (\frac{f(b)-f(a)}{b - a}=f^{\prime}(c)) for some (c\in(a,b)).

For the first - part, (f(x)=\frac{2}{2x - 1}), (a = 0), (b = 3). First, check the continuity and differentiability. The function (f(x)=\frac{2}{2x - 1}) is not continuous at (x=\frac{1}{2}\in(0,3)) (since the denominator is zero at (x = \frac{1}{2})). So, the Mean Value Theorem does not apply.

For the second - part (Question 3): By the Mean Value Theorem, (f(x)-f(a)=f^{\prime}(c)(x - a)) for some (c\in(a,x)) (or (c\in(x,a))).

  1. Let (a = 1) and (x = 7). Then (f(7)-f(1)=f^{\prime}(c)(7 - 1)) for some (c\in(1,7)). Since (f^{\prime}(x)\geq3), we have (f(7)-f(1)\geq3\times(7 - 1)). Given (f(1)=5), then (f(7)-5\geq18).
  2. Let (a = 1) and (x=-2). Then (f(-2)-f(1)=f^{\prime}(c)(-2 - 1)) for some (c\in(-2,1)). Since (f^{\prime}(x)\geq3), we have (f(-2)-f(1)\leq3\times(-2 - 1)) (because (x=-2,a = 1,x - a=-3)). Given (f(1)=5), then (f(-2)-5\leq - 9).

Step2: Solve for (f(7)) and (f(-2))

  1. For (f(7)): [ \begin{align*} f(7)-5&\geq18\ f(7)&\geq5 + 18\ f(7)&\geq23 \end{align*} ]
  2. For (f(-2)): [ \begin{align*} f(-2)-5&\leq-9\ f(-2)&\leq5-9\ f(-2)&\leq - 4 \end{align*} ]

Answer:

  1. (23)
  2. (-4)