Question

Determine whether Rolle's Theorem applies to the functionon the interval. If it applies, find all possible values of c as in the conclusion

Answer 1

The given function

`k(x)=(5x)/(x^2-9)`

on the interval [-7, 7]

Rolle's theorem : If a real valued function is countinous on [a,b], differentiable on (a,b) and f(a) = f(b) then, there exist a point c such that c belongs to (a, b) such that f'(c) = 0

Here;

`k(x)=(5x)/(x^2-9)`

Function is not countinous at x = +3, -3 which belongs [-7, 7]

Differentiate the function;

`\begin{gathered} k(x)=(5x)/(x^2-9) \\ k^(\prime)(x)=(5(-x^2-9))/((x^2-9)^2) \end{gathered}`

Here, k'(x) doesnot exists at x = +3, -3 which belong to (-7,7)

Thus, k'(x) is not differentiable on (-7, 7)

Now, for k(a) and k(b);

`\begin{gathered} k(7)=(5*7)/(7^2-9) \\ k(7)=(35)/(49-9) \\ k(7)=(35)/(40) \\ k(7)=(7)/(8) \\ \text{Now, for k=-7} \\ k(-7)=(5*(-7))/((-7)^2-9) \\ k(-7)=(-35)/(49-9) \\ k(-7)=(-35)/(40) \\ k(-7)=(-7)/(8) \\ so,\text{ k(-7)}\\e k(7) \end{gathered}`

Hence, rolle theorem doen't apply

because it is not countinous, not differentiable and k(-7) is not equal to k(7)

Answer : D,

No rolles theorem doesnot apply because k(x) is not countinous on [-7,7], not differentiable on (-7,7) and k(-7) ≠ k(7)

.......