What are the global and local extrema of #f(x)=2x^7-2x^5 # ?

Question

What are the global and local extrema of #f(x)=2x^7-2x^5  # ?

Samantha Adkison · Accepted Answer

We rewrite f as

#f(x)=2x^7*(1-1/x^2)#

but #lim_(x->oo) f(x)=oo# hence there is no global extrema.

For the local extrema we find the points where #(df)/dx=0#

#f'(x)=0=>14x^6-10x^4=0=>2*x^4*(7*x^2-5)=0=>x_1=sqrt(5/7) and x_2=-sqrt(5/7)#

Hence we have that

local maximum at #x=-sqrt(5/7)# is #f(-sqrt(5/7))=100/343*sqrt(5/7)#

and

local minimum at #x=sqrt(5/7)# is #f(sqrt(5/7))=-100/343*sqrt(5/7)#

Luke Alvarez · Answer

undefined To find the global and local extrema of the function $ f(x) = 2x^7 - 2x^5 $, we first need to find its critical points. Critical points occur where the derivative of the function is zero or undefined. Then, we evaluate the function at these critical points and at the endpoints of the domain to determine the extrema.

First, we find the derivative of $ f(x) $ with respect to $ x $:
$$ f'(x) = 14x^6 - 10x^4 $$

Next, we find the critical points by setting $ f'(x) = 0 $ and solving for $ x $:
$$ 14x^6 - 10x^4 = 0 $$
$$ 2x^4(7x^2 - 5) = 0 $$

This equation yields critical points at $ x = 0 $ and $ x = \pm \sqrt{\frac{5}{7}} $.

Now, we evaluate $ f(x) $ at these critical points and at the endpoints of the domain. The function $ f(x) $ approaches negative infinity as $ x $ approaches negative infinity and positive infinity as $ x $ approaches positive infinity. Therefore, we only need to consider the critical points.

$$ f(0) = 0 $$
$$ f\left(-\sqrt{\frac{5}{7}}\right) \approx 2.223 $$
$$ f\left(\sqrt{\frac{5}{7}}\right) \approx -2.223 $$

Thus, the global maximum is approximately $ f\left(-\sqrt{\frac{5}{7}}\right) $ and the global minimum is approximately $ f\left(\sqrt{\frac{5}{7}}\right) $. Since the function changes sign at $ x = 0 $, it has a local maximum at $ x = 0 $.