Use the function #f(x) = x^2-4sqrtx# to answer the following questions. A) Is the graph symmetric about the x or y axis? Justify your answer B) Find the asymptotes if any C) Where does the function increase and decrease?

1 Answer
Alan N.
Aug 2, 2017

A) #f(x)# is not symmetric about either #x# or #y# axes
B) #f(x)# has no asymptotes
C) #f(x)# decreases for #x in[0,1)# and increases for #x in(1,oo)#

Explanation:

#f(x) = x^2 -4sqrtx#

A) Since #sqrtx# real for #x>=0 -> f(x)# is defined for #x>=0# for #f(x) in RR#

Hence the domain of #fx) = [0, +oo)#

#:. f(x)# is undefined for #x<0# and hence is not symmetric about the #y-#axis

Now let's find the zeros of #f(x)#

#f(x) = 0 -> x^2-4sqrtx =0#

#x^2 = 4sqrtx#

#x^4 = 16x#

#x(x^3-16) =0#

#:. x = 0 or x approx 2.52#

Hence: #f(x) <=0# for #x in [0, approx 2.52]#

#f(x) > 0# for #x in (approx 2.52, +oo)#

Hence, #f(x)# cannot be symmetric about the #x-#axis.

This can be seen from the graph of #f(x)# below:

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

B) As can be seen on the graph #f(x)# has no asymtotes

C) Now let's find #f_min#

#d/dx (x^2-4sqrtx) =0#

#2x- 4. 1/2x^(-1/2) = 0#

#x-1/sqrtx =0#

#x = 1/sqrtx#

#x^2 = 1/x#

#x^3= 1 -> x=1# for #x in RR#

By observation we can see that #f(1) = f_min#

#f_min = 1^2-4sqrt1 = -3#

Hence: #f(x)# decreases for #x in[0,1)# and increases for #x in(1,oo)#

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