+118608 f(x)= 
Find the turning points of the function f'(x)=0
$$\\f(x)=2x^4+8x^{-4}\\\\
f'(x)=8x^3-32x^{-5}\\\\
f'(x)=8x^3(1-4x^{-8})\\\\
f"(x)=24x^2+160x^{-6}>0 $ except when x=0 but $x\ne 0\\\\
x\ne 0 \;\;$so minimum points occur when $\\\\
1-\;4x^{-8}=0\\\\
1=\;4x^{-8}\\\\
\frac{1}{4}=\frac{1}{x^8}\\\\
x^8=4\\\\
x=\pm\;2^{1/4}\\\\
x\approx \pm\;1.1892\\\\$$
$$\\When\;\; x=\pm\;2^{1/4}\\\\
f(\pm 2^{1/4})=2\times (2^{1/4})^4+8(2^{1/4})^{-4}\\\\
f(\pm 2^{1/4})=2\times 2+8(2^{-1})\\\\
f(\pm 2^{1/4})=4+4\\\\
f(\pm 2^{1/4})=8\\\\$$
So the smallest value of the expression is 8
Here is a graph to verify what i have said 
+128406 So we want to minimize 2x^4 + 8x^(-4)
Take the derivative and set this to 0
8x^3 - 32x^(-5) = 0 factor
8x^(-5) [x^8 - 4] = 0 ignore the frst factor, it can never be 0
x^8 - 4 = 0
x^8 = 4 take the ±roots.... x = ±(4)^(1/8) = about ±1.189
Now, take the second derivative and see which critical points (if any) produce a positive result
24x^2 + 160x^(-6) and both points will produce positive results, so both are minimums
Here's the graph of the function........https://www.desmos.com/calculator/oeqrmz5jb1
+118608 f(x)=
Find the turning points of the function f'(x)=0
$$\\f(x)=2x^4+8x^{-4}\\\\
f'(x)=8x^3-32x^{-5}\\\\
f'(x)=8x^3(1-4x^{-8})\\\\
f"(x)=24x^2+160x^{-6}>0 $ except when x=0 but $x\ne 0\\\\
x\ne 0 \;\;$so minimum points occur when $\\\\
1-\;4x^{-8}=0\\\\
1=\;4x^{-8}\\\\
\frac{1}{4}=\frac{1}{x^8}\\\\
x^8=4\\\\
x=\pm\;2^{1/4}\\\\
x\approx \pm\;1.1892\\\\$$
$$\\When\;\; x=\pm\;2^{1/4}\\\\
f(\pm 2^{1/4})=2\times (2^{1/4})^4+8(2^{1/4})^{-4}\\\\
f(\pm 2^{1/4})=2\times 2+8(2^{-1})\\\\
f(\pm 2^{1/4})=4+4\\\\
f(\pm 2^{1/4})=8\\\\$$
So the smallest value of the expression is 8
Here is a graph to verify what i have said
