setting the derivative equaled to zero to find critical points
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Adding 1/5 to each side: x/3(25+x^2)^-1/2=1/5.
Squaring and cross-multiplying: 25x^2=9(25+x^2)
Expanding: 25x^2=225+9x^2
Solving for x: 16x^2=225; x=15/4=3.75 (only the positive root is valid in original equation).
Derivative of original function (differentiating by parts): 1/3(25+x^2)^-1/2-x^2(25+x^2)^-3/2=1/3(25+x^2)^-1/2(1-x^2/(25+x^2))
=1/3(25+x^2)^-1/2((25+x^2-x^2)/(25+x^2))
=25/3(25+x^2)^-3/2.
This derivative is always positive for all real values of x, and at x=3.75 has the value 64/1875=0.03413.
The nature of the derivative at x=3.75 is neither a maximum nor a minimum.
The root of the original equation, x=3.75, is a point of inflection. At 3.75- the function is negative and at 3.75+ it is positive (where - and + denote just less and just greater than).
by Top Rated User (1.2m points)
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