Question #54b57

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Answer 1 · 2018-01-24T04:50:49

#(5/12, 13/12)#

#f(x)=sqrt(3x+1) - x#

#f(x)= (3x + 1)^(1/2) - x#

Take the first derivative:

#f'(x) = (1/2)(3x + 1)^(-1/2)*3 - 1#

#f'(x)=3/(2sqrt(3x+ 1)) - 1#

Set the first derivative to zero to solve for critical point(s):

#3/(2sqrt(3x+ 1))-1=0#

#(3 - 2sqrt(3x+1))/(2sqrt(3x+1))=0# => denominator can't be zero:

#3=2sqrt(3x+1)# => square both sides:

#9=4(3x+1)#

#9=12x+4#

#5=12x#

#x=5/12# => find the y value at this point:

#f(5/12)=sqrt(5/4+1)-5/12=sqrt(9/4)-5/12#

#=3/2-5/12=18/12-5/12=13/12#

Thus our critical point is: #(5/12, 13/12)# now to determine the

nature of this point we use the 2nd derivative test:

#f ''(x)= -9/(4 (3 x + 1)^(3/2))#

#f ''(5/12)=-2/3<0#

This proves that our critical point is indeed a global/absolute

maximum.