`L e tf(x)={x e^(a x),xlt=0x+a x^2-x^3,x >0` where `a` is a positive constant. Find the interval in which `f^(prime)(x)` is increasing.

Explanation

At ` x =0, LHL = underset(x to 0^(-))(lim) f(x) = underset (x to 0^(-)) (lim) x e ^(ax ) = 0`
and `RHL = underset (x to 0^(+ )) (lim) f(x) = underset (x to 0^(+)) ( lim) (x + ax^(2) - x ^(3)) = 0`
Therefore, `LHL = RHL =0 = f (0) `
So, `f(x)` is continuous at x=0
Also, ` f' (x) = {{:(1*e^(ax) + ax e ^(ax)",",,if x lt 0),(1+ 2ax - 3x ^(2) ",",, if x gt 0):}`
and ` Lf' (0) = underset (x to 0^(-))(lim) ( f(x) - f(0))/( x - 0)`
`" " = underset (x to 0 ^(-)) ( xe^(ax) - 0 ) /( x ) = underset (x to 0^(-)) (lim) e ^(ax) = e ^(0) = 1 `
and ` Rf' (0) = underset (x to 0^(+)) (lim) (f (x) - f(0))/( x + 0)`
`" " = underset(x to 0^(+)) (lim) (x + a x^(2) - x ^(3)-0)/(x) `
`" " = underset ( x to 0^(+)) (lim) 1+ ax - x ^(2) =1 `
Therefore, `Lf'(0) = Rf' (0) =1 rArr f ' (0) = 1 `
Hence, `f' (x) = {{:((ax + 1)e ^(ax)",",,if x lt 0), (1",",, if x =0), (1+ 2 ax - 3x ^(2)",",, if x gt 0):}`
Now, we can say without solving that, ` f' (x)` is continuous at ` x =0` and hence on R. We have,
`f ''(x) = {{:(ae^(ax)+ a (ax + 1 ) e^(ax) "," ,, if x lt 0 ), ( 2a - 6x ",",, if x gt 0):}`
and `Lf'' (0)= underset(x to 0^(-)) (lim) (f' (x) - f' (0))/(x - 0)`
`" " = underset(x to 0^(-)) (lim) ((ax + 1) e ^(ax) - 1)/( x)`
`" " = underset (xto 0^(-)) (lim) [ ae ^(ax) + (e ^(ax) - 1)/( x ) ]`
`= underset (x to 0 ^(-)) (lim) ae ^(ax) + a * underset (x to 0 ^(-)) (lim) ( e ^(ax ) - 1)/( ax) `
`" " = ae^(0) + a (1) = 2a `
and ` R f'' (0) = underset (x to 0 ^(+)) (lim) ( f ' (x) - f ' (0)) /(x + 0) `
`" " = underset (x to 0^(+)) (lim) ((1 + 2 ax - 3x ^(2)) - 1)/( x )`
`" " = underset (x to 0^(+)) (lim) ( 2ax - 3x^(2) ) /(x) = underset ( x to 0^(+)) (lim) 2 a -3x = 2a `
Therefore, ` Lf '' (0) = Rf'' (0) = 2a `
Henc, ` f'' (x ) = {{:(a (ax + 2) e^(ax)",",, if x lt 0 ), ( 2 a",",, if x =0), (2a - 6x ",",, if x gt 0 ):}`
Now, for ` x lt 0, f' ' (x) gt 0, if ax + 2 gt 0`
`rArr "For " x lt 0, f '' (x) gt 0 , if x gt - 2 //a`
`rArr " " f ' (x) gt 0 , if -(2)/(a) lt x lt 0`
and for ` x gt 0, f '' (x) gt 0, if 2a - 6x gt 0`
`rArr " for " x gt 0, f'' (x) gt 0, if x lt a //3 `
Thus, `f(x)` is increasing.
When `" " x ge 0, f (x) ge f (0)`
`rArr " " 2 sin x + 2tan x - 3 x gt 0 + 0 -0 `
`rArr " " 2 sin x + 2 tan x - 3x ge 3x `