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

$k(x)=\frac{5x}{x^2-9}$

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;

$k(x)=\frac{5x}{x^2-9}$

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

Differentiate the function;

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

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);

$\begin{gathered} k(7)=\frac{5\times7}{7^2-9} \\ k(7)=\frac{35}{49-9} \\ k(7)=\frac{35}{40} \\ k(7)=\frac{7}{8} \\ \text{Now, for k=-7} \\ k(-7)=\frac{5\times(-7)}{(-7)^2-9} \\ k(-7)=\frac{-35}{49-9} \\ k(-7)=\frac{-35}{40} \\ k(-7)=\frac{-7}{8} \\ so,\text{ k(-7)}\ne k(7) \end{gathered}$

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)

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