What are the extrema of # f(x)=1/x^3 +10x# on the interval [1,6]?

Question

Ezra Adams · Accepted Answer

Always start with a sketch of the function over the interval.

On the interval [1,6], the graph looks like this:

As observed from the graph, the function is increasing from 1 to 6. So, there is no local minimum or maximum .

However, the absolute extrema will exist at the endpoints of the interval:

absolute minimum : f(1) #= 11#
absolute maximum : f(6) #=1/216+60~~60.005#

hope that helped

Christopher Allers · Answer

undefined To find the extrema of $ f(x) = \frac{1}{x^3} + 10x $ on the interval $[1,6]$, we need to find critical points and then evaluate $ f(x) $ at those points as well as at the endpoints of the interval.

1. Find the derivative of $ f(x) $ with respect to $ x $:
$$ f'(x) = -\frac{3}{x^4} + 10 $$

2. Set $ f'(x) $ equal to zero and solve for $ x $ to find the critical points:
$$ -\frac{3}{x^4} + 10 = 0 $$
$$ \frac{3}{x^4} = 10 $$
$$ x^4 = \frac{3}{10} $$
$$ x = \sqrt[4]{\frac{3}{10}} $$

3. Evaluate $ f(x) $ at the critical point and the endpoints of the interval:
$$ f(1) = 1 + 10 = 11 $$
$$ f(6) = \frac{1}{216} + 60 = \frac{61}{216} $$
$$ f(\sqrt[4]{\frac{3}{10}}) = \frac{10}{3}\sqrt[4]{\frac{3}{10}} $$

4. Compare the values obtained:
   - The critical point is within the interval $[1,6]$, so we consider $ f(\sqrt[4]{\frac{3}{10}}) $.
   - Compare $ f(1) $, $ f(6) $, and $ f(\sqrt[4]{\frac{3}{10}}) $ to determine which one is the minimum and which one is the maximum.

So, the minimum and maximum values occur at the critical point $ \sqrt[4]{\frac{3}{10}} $ and the endpoint $ x = 6 $, respectively.