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

Explanation:

Step1: Calculate (f(3)) and (f(0))

• For (f(x)=\frac{2}{2x - 1}), when (x = 3), (f(3)=\frac{2}{2\times3-1}=\frac{2}{5}).
• When (x = 0), (f(0)=\frac{2}{2\times0 - 1}=- 2).
• Then (\frac{f(3)-f(0)}{3 - 0}=\frac{\frac{2}{5}-(-2)}{3}=\frac{\frac{2 + 10}{5}}{3}=\frac{\frac{12}{5}}{3}=\frac{4}{5}).

Step2: Find (f^{\prime}(x))

• Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 2), (u^\prime=0), (v=2x - 1), (v^\prime = 2).
• (f^{\prime}(x)=\frac{0\times(2x - 1)-2\times2}{(2x - 1)^{2}}=\frac{-4}{(2x - 1)^{2}}).

Step3: Solve for (c)

• Set (f^{\prime}(c)=\frac{4}{5}), so (\frac{-4}{(2c - 1)^{2}}=\frac{4}{5}).
• Cross - multiply gives (-20 = 4(2c - 1)^{2}).
• Then ((2c - 1)^{2}=- 5).

Since the square of a real number ((2c - 1)^{2}\geq0) for all real (c), and (-5<0), there is no real number (c) that satisfies the equation.

Answer:

DNE