Suppose #g# is a function whose derivative is #g'(x)=3x^2+1# Is #g# increasing, decreasing, or neither at #x=0#?

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Answer 1 · 2018-05-07T20:05:53

Increasing

#g'(x)=3x^2+1>0# , #AA##x##in##RR# so #g# is increasing in #RR# and so is at #x_0=0#

Another approach,

#g'(x)=3x^2+1# #<=>#

#(g(x))'=(x^3+x)'# #<=>#

#g#, #x^3+x# are continuous in #RR# and they have equal derivatives, therefore there is #c##in##RR# with

#g(x)=x^3+x+c#,
#c##in##RR#

Supposed #x_1#,#x_2##in##RR# with #x_1<##x_2# #(1)#

#x_1<##x_2# #=># #x_1^3<##x_2^3# #=># #x_1^3+c<##x_2^3+c# #(2)#

From #(1)+(2)#

#x_1^3+x_1+c<##x_2^3+x_2+c# #<=>#

#g(x_1)<##g(x_2)# #-># #g# increasing in #RR# and so at #x_0=0##in##RR#