Given:
The function
![f(x)=(x)/(x-5),\text{ on interval}[1,4]](https://img.qammunity.org/2023/formulas/mathematics/college/e3mprj41gycsvao5vk6ligxv53k4pg3wbn.png)
Required:
Dermine whether the Mean Value Theorem can be applied to f on the closed interval [a,b]. If the Mean Value Theorem can be applied, find all values of in the open interval .
Step-by-step explanation:
![\begin{gathered} \text{The mean value theorem states that for a continuous and differentiable} \\ \text{ function }f(x)\text{ on the interval }[a,b],\text{ there exists such number }c\text{ from } \\ \text{ interval }(a,b),\text{ that }f^(\prime)(c)=(f(b)-f(a))/(b-a). \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4o7vfkvsd6bljts2f01kwj47ky88pfj8bv.png)
First, evaluate the function at the endpoints of the interval:
![\begin{gathered} f(4)=-4 \\ f(1)=-(1)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d85eqmssu1qtjle9ltgrscp3g7ilbbx7g5.png)
Next, find the derivative
![\begin{gathered} f^(\prime)(c)=-(c)/((c-5)^2)+(1)/(c-5) \\ =((-4)-(-(1)/(4)))/((4)-(1)) \\ Simplify: \\ -(c)/((c-5)^2)+(1)/(c-5)=-(5)/(4) \\ Sol\text{ve the equation on the given interval}:c=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/299s99nvstmayi2540g5mr3qcvzxk5act7.png)
Answer:
The value of c = 3.