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(c) )? if there is, enter your answer below (write dne if there is no value of ( c )).

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