Question

find the absolute extrema of the function, if they exist, over the indicated interval. also indicate the x - value at which each extremum occurs. if no interval is specified, use the real numbers. (-∞, ∞)
n(x)=\sqrt7{x}, 0,128
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice
a. the absolute maximum is at x = and the absolute minimum is at x =
(use a comma to separate answers as needed )
b. the absolute minimum is at x = and there is no absolute maximum
(use a comma to separate answers as needed )
c. the absolute maximum is at x = and there is no absolute minimum
(use a comma to separate answers as needed )
d. there is no absolute maximum and no absolute minimum

Explanation:

Step1: Find the derivative of the function

The function is \(N(x)=\sqrt[7]{x}=x^{\frac{1}{7}}\).
Using the power rule \(y = x^n\), \(y^\prime=nx^{n - 1}\), we have \(N^\prime(x)=\frac{1}{7}x^{\frac{1}{7}-1}=\frac{1}{7}x^{-\frac{6}{7}}=\frac{1}{7x^{\frac{6}{7}}}\).
The derivative \(N^\prime(x)\) is never zero (since the numerator is non - zero for all \(x\)) and is undefined at \(x = 0\). But \(x = 0\) is an endpoint of the interval \([0,128]\).

Step2: Evaluate the function at the endpoints

Evaluate \(N(x)\) at \(x = 0\) and \(x=128\).
When \(x = 0\), \(N(0)=\sqrt[7]{0}=0\).
When \(x = 128\), \(N(128)=\sqrt[7]{128}\). Since \(128 = 2^7\), then \(N(128)=\sqrt[7]{2^7}=2\).

Answer:

A. The absolute maximum is \(2\) at \(x = 128\) and the absolute minimum is \(0\) at \(x=0\)