Question 1027080
Use the derivative,
{{{f(x)=2x^4+8x^(-4)}}}
{{{df/dx=8x^3-32x^(-5)}}}
Set it equal to zero,
{{{8x^3-32x^(-5)=0}}}
{{{(8x^8-32)/x^5=0}}}
{{{8x^8=32}}}
{{{x^8=1/4}}}
{{{x=0 +- (4)^(1/8)}}}

{{{x=0 +- (2)^(1/4)}}}
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.
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So then,
{{{f[max]=2*(2^(1/4))^4+8/((2^(1/4))^4)}}}
{{{f[max]=2*2+8/2}}}
{{{f[max]=4+4}}}
{{{f[max]=8}}}

*[illustration dx5.JPG].