Question

Test `f` for concavity?

`f` 2 times differentiable in `RR` with:

`(f'(x))^3+3f'(x)=e^x+cosx+x^3+2x+7`

`AA` `x` `in` `RR`

Answer 1

`f` is convex in `RR`

Explanation:

Solved it i think.

`f` is 2 times differentiable in `RR` so `f` and `f'` are continuous in `RR`

We have `(f'(x))^3+3f'(x)=e^x+cosx+x^3+2x+7`

Differentiating both parts we get

`3*(f'(x))^2f''(x)+3f''(x)=e^x-sinx+3x^2+2` `<=>`

`3f''(x)((f'(x))^2+1)=e^x-sinx+3x^2+2`

• `f'(x)^2>=0` so `f'(x)^2+1>0`

`<=>` `f''(x)=(e^x-sinx+3x^2+2)/(3((f'(x))^2+1)>0)`

We need the sign of the numerator so we consider a new function

`g(x)=e^x-sinx+3x^2+2`
, `x` `in` `RR`

`g'(x)=e^x-cosx+6x`

We notice that `g'(0)=e^0-cos0+6*0=1-1+0=0`

For `x=π` `=>` `g'(π)=e^π-cosπ+6π=e^π+1+6π>0`

For `x=-π` `g'(-π)=e^(-π)-cos(-π)-6π=1/e^π+cosπ-6π=1/e^π-1-6π<0`

We finally get this table which shows the monotony of `g`

Supposed `I_1=(-oo,0]` and `I_2=[0,+oo)`
`g(I_1)=g((-oo,0])=[g(0),lim_(xrarr-oo)g(x))=[3,+oo)`

`g(I_2)=g([0,+oo))=[g(0),lim_(xrarr+oo)g(x))=[3,+oo)`

because

• `lim_(xrarr-oo)g(x)=lim_(xrarr-oo)(e^x-sinx+3x^2+2)`

`|sinx|<=1` `<=>` `-1<=-sinx<=1` `<=>`

`e^x+3x^2+2-1<=e^x+3x^2+2-sinx<=e^x+3x^2+2+1` `<=>`

`e^x+3x^2+1<=e^x-sinx+3x^2+2<=e^x+3x^2+3 <=>`

`e^x+3x^2+1<=g(x)<=e^x+3x^2+3`

• Using the squeeze/sandwich theorem we have

`lim_(xrarr-oo)(e^x+3x^2+1)=+oo=lim_(xrarr-oo)(e^x+3x^2+3x)`

Therefore, `lim_(xrarr-oo)g(x)=+oo`

• `lim_(xrarr+oo)g(x)=lim_(xrarr+oo)(e^x-sinx+3x^2+2)`

With the same process we end up to

`e^x+3x^2+1<=g(x)<=e^x+3x^2+3`

However, `lim_(xrarr+oo)(e^x+3x^2+1)=+oo=e^x+3x^2+3`

Therefore, `lim_(xrarr+oo)g(x)=+oo`

The range of `g` will be:

`R_g=g(D_g)=g(I_1)uug(I_2)=[3,+oo)`

• `0!inR_g=[3,+oo)` so `g` has no roots in `RR`
  `g` is continuous in `RR` and has no solutions. Therefore, `g` preserves sign in `RR`

That means

`{(g(x)>0" , "xεRR),(g(x)<0" , "xεRR):}`

Thus, `g(π)=e^π-sinπ+3π^2+2=e^π+3π^2+2>0`

As a result `g(x)>0`, `x` `in` `RR`

And `f''(x)>0` , `x` `in` `RR`

`->` `f` is convex in `RR`

Answer 2

See below.

Explanation:

Given `y = f(x)` the curve curvature radius is given by

`rho = (1+(f')^2)^(3/2)/(f'')` so given

`(f')^3+3f' = e^x + cosx + x^3 + 2 x + 7` we have

`3(f')^2f''+3f'' = e^x+3x^3-sinx+2` or

`f''(1+(f')^2) = 1/3( e^x+3x^3-sinx+2)` or

`1/(f''(1+(f')^2))=3/(e^x+3x^3-sinx+2)` or

`rho = (1+(f')^2)^(3/2)/(f'') = (3(1+(f')^2)^(5/2))/(e^x+3x^3-sinx+2)`

now analyzing `g(x) = e^x+3x^3-sinx+2` we have

`min g(x) = 0` for `x in RR` so `g(x) ge 0` and then the curvature in

`rho = (3(1+(f')^2)^(5/2))/(e^x+3x^3-sinx+2)` doesn't changes sign so we conclude that `f(x)` epigraph is convex in `RR`