Question

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 )).
question 3
let ( 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 ).

1. what is the smallest possible value for ( f(7) )?

answer:

1. what is the largest possible value for ( f(-2) )?

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 \(f(7)\):

$$
LATEXBLOCK0
$$

1. For \(f(-2)\):

$$
LATEXBLOCK1
$$

Answer:

1. \(23\)
2. \(-4\)