Use the function f(x) = x^2-4sqrtx to answer the following: A) Where is the curve concave up or down? B) Locate the points of inflection (if any). Help!?

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Answer 1 · 2017-08-02T06:02:20

The curve is concave up for #x in (0,+oo)# and there are no points of inflections

The function is #f(x)=x^2-4sqrtx#

The domain of #f(x)# is #x in [0,+oo)#

The first derivative of #f(x)# is

#f'(x)=2x-(4)/(2sqrtx)=2x-2x^(-1/2)#

#f'(x)=0#, #=>#, #2x-2/sqrtx=0#

#xsqrtx=1#

Squaring both sides

#x^3=1#

#x=1# is a critical point

The second derivative is

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

#f''(0)#, #=>#, #2+1/x^(3/2)=0#

#2x^(3/2)+1=0#

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

The solution to this equation is #S=O/#

When #x=1#

#f''(1)=3#

As #f''(1)>0#, this is a minimum point and the curve is concave up

There are no points of inflections

graph{x^2-4sqrtx [-10, 10, -5, 5]}