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
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Let y = (1/3)x^2 = g(x)
We are looking for the point where y = x + c is tangent to g(x)
Beause the slope of y = x + c = 1 we are looking for the point where the derivative of g(x) has a slope = 1.
So
g ' (x) = 2/3x ....set this = 1 and x = 3/2
And to find the y coordinate of this point we have y= (1/3)(3/2)^2 = (1/3)(9/4) = 9/12 = 3/4
So, since the point (3/2, 3/4) is also on y = x + c we have
3/4 = 3/2 + c subtract 3/2 from both sides
c = -3/4
Here's the graph.....https://www.desmos.com/calculator/ugbqgcgqrf
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