Melody

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यूजर का नामMelody
स्कोर118733
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Melody  11 Feb 2022
 #3
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1 Nov 2014
 #2
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+5

f(x)=   2x^4 + \frac{8}{x^4} 

 

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  

 

https://www.desmos.com/calculator/2ybonktd3t

1 Nov 2014