How do you find the exact relative maximum and minimum of the function of #f(x) = 4x+6x^-1#?

Question

Charles Arocha · Accepted Answer

Find the critical numbers. Test the critical numbers. Evaluate #f# at the critical numbers.

#f(x) = 4x+6x^-1# has domain #(-oo,0)uu(0,oo)# #f'(x) = 4-6/x^2# is undefined only at #x=0# (which is not in the domain of #f#, hence is not a critical number). #f'(x) = 0# at #x=+-sqrt(3/2)# This is a good example to use the second derivative test for. #f''(x) = 12/x^3# #" " # (which is continuous except at #x=0#) So #f''(-sqrt(3/2)) < 0#, which implies that #f(-sqrt(3/2))# is a relative maximum. And #f''(sqrt(3/2)) > 0#, which implies that #f(sqrt(3/2))# is a relative minimum. We can save ourseles some work evaluating the function by noting that the function is odd. So #f(-sqrt(3/2)) = f(sqrt(3/2))#. #f(sqrt(3/2)) = (4sqrt3)/sqrt2 + (6sqrt2)/sqrt3# # = 2sqrt2sqrt3+2sqrt3sqrt2# # = 4sqrt6#. #f# has a relative maximum of #-4sqrt6# (at #x=-sqrt(3/2)# and a relative minimum of #4sqrt6# (at #x=sqrt(3/2)#.

Aurora Alvarado · Answer

undefined To find the exact relative maximum and minimum of the function $ f(x) = 4x + 6x^{-1} $, follow these steps:

1. Find the derivative of the function $ f'(x) $.
2. Set $ f'(x) $ equal to zero and solve for $ x $ to find critical points.
3. Use the second derivative test or analyze the behavior of $ f'(x) $ around the critical points to determine the nature of the extremum (maximum or minimum).

Let's go through these steps:

1. Find the derivative of the function $ f(x) $:
$$ f'(x) = 4 - 6x^{-2} $$

2. Set $ f'(x) $ equal to zero and solve for $ x $ to find critical points:
$$ 4 - 6x^{-2} = 0 $$
$$ 6x^{-2} = 4 $$
$$ x^{-2} = \frac{4}{6} = \frac{2}{3} $$
$$ x^2 = \frac{3}{2} $$
$$ x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2} $$

So, the critical points are $ x = \frac{\sqrt{6}}{2} $ and $ x = -\frac{\sqrt{6}}{2} $.

3. Determine the nature of the extremum using the second derivative test:
$$ f''(x) = 12x^{-3} $$

Evaluate $ f''(x) $ at the critical points:
- At $ x = \frac{\sqrt{6}}{2} $, $ f''\left(\frac{\sqrt{6}}{2}\right) = 12\left(\frac{\sqrt{6}}{2}\right)^{-3} = 12\sqrt{6} $, which is positive. So, there is a relative minimum at $ x = \frac{\sqrt{6}}{2} $.
- At $ x = -\frac{\sqrt{6}}{2} $, $ f''\left(-\frac{\sqrt{6}}{2}\right) = 12\left(-\frac{\sqrt{6}}{2}\right)^{-3} = -12\sqrt{6} $, which is negative. So, there is a relative maximum at $ x = -\frac{\sqrt{6}}{2} $.

Therefore, the exact relative minimum occurs at $ x = \frac{\sqrt{6}}{2} $ and the exact relative maximum occurs at $ x = -\frac{\sqrt{6}}{2} $.