Question

In Exercises \(11-18,\) use analytic methods to find the extreme values of the function on the interval and where they occur. $$f(x)=x^{2 / 5}, \quad-3 \leq x<1$$

Short answer

The minimal value of the function on the given interval is 0 at \( x = 0 \), and the maximal value is 1.24573 at \( x = -3 \).

Step-by-step solution

Identify the Function and the Interval

We are dealing with the function \( f(x) = x^{2 / 5} \) and the interval \(-3 \leq x < 1\). It's clearly defined.

Compute the Derivative of the Function

The derivative of this function, \( f'(x) \), is found by applying the power rule for differentiation to the function: \( f'(x) = (2 / 5)x^{-3 / 5}\).

Find the Critical Points

Critical points occur where the derivative is zero or undefined. However, \( f'(x) = (2 / 5)x^{-3 / 5} \) is never zero and is undefined at \( x = 0 \). So, 0 is the only critical point.

Evaluate the Function at the Critical Points and Endpoints

We evaluate the function at the critical point 0 and the endpoints -3 and 1 to find possible extreme values. We get: \( f(-3) = (-3)^{2 / 5} = 1.24573, f(0) = 0^{2 / 5} = 0, and f(1) is not defined because 1 is not included in the interval.

Identify the Extreme Values

The largest value obtained is the maximum value and the smallest value is the minimum value. The minimum value of \( f(x) \) on the interval \(-3 \leq x < 1\) is 0, which occurs at \( x = 0 \). The maximum value is 1.24573, which occurs at \( x = -3 \).