How do you find the critical numbers for #f(x)= x sqrt (2x+1)# to determine the maximum and minimum?

1 Answer
Apr 28, 2017

Answer:

Critical points are #x=-1/2#, where #f'(x)# is not defined and #x=-1/3#, where we have a minima for #f(x)#

Explanation:

We say that #x=a# is a critical point of the function #f(x)# if #f(a)# exists and if either #f'(a)=0# or #f'(a)# does not exist.

Observe that #f(x)=xsqrt(2x+1)# if #x < -1/2#

Let us therefore workout #f'(x)# using product rule.

#f'(x)=x xx1/(2sqrt(2x+1))xx2+1xxsqrt(2x+1)#

= #x/(sqrt(2x+1))+sqrt(2x+1)#

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

Now #f'(x)=0#, when #x=-1/3#. Also #f'(x)# is not defined for #x <= -1/2#, hence #x=-1/2# is a critical point.

Further #f''(x)=(sqrt(2x+1)xx3-(3x+1)xx(1/sqrt(2x+1)))/(2x+1)#

= #(6x+3-3x-1)/((2x+1)sqrt(2x+1))=(3x+2)/((2x+1)sqrt(2x+1)#

and at #x=-1/3#, #f''(x) > 0# and hence we have a minima at #x=-1/3#, where #f(x)=-1/(3sqrt3)#.

graph{xsqrt(2x+1) [-4.08, 5.92, -0.86, 4.14]}