How do you find the extrema for #g(x) = sqrt(x^2 + 2x + 5)#?

2 Answers
May 4, 2018

Answer:

#g(x)# has no maximum and a global and local minimum in #x=-1#

Explanation:

Note that:

#(1) " "x^2+2x+5 = x^2+2x+1+4 = (x+1)^2+4 > 0#

So the function

#g(x) = sqrt(x^2+2x+5)#

is defined for every #x in RR#.

Besides as #f(y) = sqrty# is a monotone increasing function, then any extremum for #g(x)# is also an extremum for:

#f(x) = x^2+2x+5#

But this is a second order polynomial with leading positive coefficient, hence it has no maximum and a single local minimum.

From #(1)# we can easily see that as:

#(x+1)^2 >= 0#

and:

#x+1=0#

only when #x=-1#, then:

#f(x) >= 4#

and

#f(x) = 4#

only for #x=-1#.

Consequently:

#g(x) >= 2#

and:

#g(x) = 2#

only for #x=-1#.

We can conclude that #g(x)# has no maximum and a global and local minimum in #x=-1#

May 4, 2018

#g(x)=sqrt(x^2+2x+5)# , #x##in##RR#

We need #x^2+2x+5>=0#

#Δ=2^2-4*1*5=-16<0#

#D_g=RR#

#AA##x##in##RR#:

#g'(x)=((x^2+2x+5)')/(2sqrt(x^2+2x+5))# #=#

#(2x+2)/(2sqrt(x^2+2x+5))# #=#

#(x+1)/(sqrt(x^2+2x+5)>0)#

#g'(x)=0# #<=># #(x=-1)#

  • For #x<-1# we have #g'(x)<0# so #g# is strictly decreasing in #(-oo,-1]#

  • For #x>##-1# we have #g'(x)>0# so #g# is strictly increasing in #[-1,+oo)#

Hence #g(x)>=g(-1)=2>0# , #AA##x##in##RR#

As a result #g# has a global minimum at #x_0=-1# , #g(-1)=2#

enter image source here