Simplify the expression so there is only one positive power for each base
1 answer:
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One way to do it is with calculus. The distance between any point 
on the line to the origin is given by
Now, both
and 
attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.
Solving for
, you find a critical point of 
.
Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.
You have
so indeed, a minimum occurs at.
The minimum distance is then
Now, both
Solving for
Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.
You have
so indeed, a minimum occurs at
The minimum distance is then