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#

2 Answers
Jan 16, 2018

#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#
enter image source here

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)#

enter image source here

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#

Jan 16, 2018

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#