How do you determine whether the function satisfies the hypotheses of the mean value theorem for \[f(x)={{x}^{\dfrac{1}{3}}}\] on the interval \[\left[ -5,4 \right]\]?

Hint: The mean value theorem holds true for a function f on range \[\left[ a,b \right]\], when it satisfies two conditions or hypotheses given below: The first condition is that the function should be continuous for \[\left[ a,b \right]\]. The second condition is that the functions should be differentiable on \[\left( a,b \right)\]. If the function satisfies these two hypotheses, the mean value theorem holds for the function. We will check if the given function satisfies these conditions or not, and then also find the constant value.

Complete step by step solution:
We are given the function \[f(x)={{x}^{\dfrac{1}{3}}}\], we have to check if it satisfies the conditions for mean value theorem on the interval \[\left[ -5,4 \right]\].
As this is an exponent function, this will be continuous on the range \[\left[ -\infty ,\infty \right]\]. This means that the function is continuous on \[\left[ -5,4 \right]\]. Thus, it satisfies the first hypotheses.
Differentiating the function, we get
\[\begin{align}
  & f'(x)=\dfrac{d\left( {{x}^{\dfrac{1}{3}}} \right)}{dx} \\
 & f'(x)=\dfrac{1}{3}{{x}^{\dfrac{-2}{3}}}=\dfrac{1}{3{{x}^{\dfrac{2}{3}}}} \\
\end{align}\]
We can see that this derivative exists for all values of x except 0. Thus, for one value of x in the given range, the function does not satisfy the second hypothesis.
Although this function does not satisfy the hypotheses on the interval, it does satisfy the conclusion. Hence, the function satisfies both hypotheses for mean value theorem.

Note: As this function satisfies the two conditions for mean value theorem, we can also find the constant of mean value theorem as follows:
\[\begin{align}
  & f'(x)=\dfrac{f(4)-f(-5)}{4-(-5)} \\
 & \dfrac{1}{3{{x}^{\dfrac{2}{3}}}}=\dfrac{({{4}^{\dfrac{1}{3}}})-{{(-5)}^{\dfrac{1}{3}}}}{9} \\
\end{align}\]
Simplifying the above equation, we get
\[x={{\left( \dfrac{1}{3\left( \dfrac{({{4}^{\dfrac{1}{3}}})-{{(-5)}^{\dfrac{1}{3}}}}{9} \right)} \right)}^{\dfrac{3}{2}}}\]
We can use a calculator to find the value of above expression, we get
\[x\approx \pm 0.87\]