Determine the point of maxima and minima of the function `f(x)=1/8(log)_e x-b x+x^2,x >0,` where `bgeq0` is constant.

Answer

Maxima at ` x = ((b - sqrt ( b ^(2) - 1 ))/( 4) ` and minima at ` x = (1)/(4) (b + sqrt (b ^(2) -1 ))`

Explanation

`f (x)`is differentiable function for ` x gt 0`.
Therefore, for maxima or minima, ` f ' (x) = 0` must satisfy.
Given, ` " " f(x) = (1) /(8) " in " x - b x + x ^(2) , x gt 0 `
`rArr " " f ' (x) = (1)/(8)*(1)/(x) - b + 2x `
For ` " " f'(x) = 0`
`rArr (1)/(8x ) - b + 2x = 0`
`rArr 16 x ^(2) - 8bx + 1 = 0`
`rArr " " ( 4x -b ) ^(2) = b^ (2)- 1" "` ... (i)
`rArr " " ( 4x - b ) ^(2) = (b - 1) (b + 1)" " [ b ge 0, `given]
Case I ` 0 le b lt 1`, has no solution. Since, RHS is negative in this domain and LHS is positive.
Case II When `b = 1`, then `x = (1)/(4 )` is the only solution.
When ` b = 1,`
` f ' (x) = (1)/(8x ) - 1 + 2 x = (2)/(x ) ( x ^(2) - (1)/(2) x + (1)/(1 6) ) = ( 2)/(x) ( x - (1)/(4) ) ^(2)`
We have to check the sign of `f ' (x)` at ` x = 1//4`

From sign chart, it is clear that ` f ' (x)` has no change of sign in left and right of x = 1/4.
Case III When ` b gt 1 ` , then
`f ' (x) = (1)/(8x) - b + 2x = ( 2)/(x) ( x ^(2) - (1)/(8) bx + (1)/(16))`
` = (2)/(x) [ (x - (b)/(4)) ^(2) - (1)/(16) (b ^(2) - 1)]`
` " " = (2)/(x) [ (x - (b)/(4) - (1)/(4) sqrt ( b ^(2) - 1)) (x - (b)/(4) + (1)/(4) sqrt (b ^(2) - 1)) ]`
` = (2)/(x) (x - alpha ) (x - beta ) `
where, ` alpha lt beta and alpha = (1)/(4) (b - sqrt (b ^(2) - 1) and `
`beta = (1)/(4) (b + sqrt (b ^(2) - 1).` From sign scheme, it is clear that
`" " f' (x){{:( gt 0",",, " for " 0 lt x lt alpha), ( lt0",",, " for " alpha lt x lt beta ) ,( gt 0",",, " for " x gt beta ):}`
By the first derivative test, `f(x)` has a maxima at ` x = alpha `
` " " = (1)/(4) (b - sqrt (b^(2) - 1))`
and ` f(x) ` has a minima at ` x = beta = (1)/(4) (b + sqrt (b^(2) - 1))`