How do you graph #f(x)=x^5+2# using zeros and end behavior?

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Answer 1 · 2017-03-19T11:20:57

#f(x)# has a real zero at #x~=- 1.148698# and an inflection point at #(0,2)#
#f(x)# is defined #forall x in RR#

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

Consider: #f(x) =0 -> x^5 = -2#

#:. x = root5(-2) = -1.148698# for #x in R#

Now consider, #f(0) = 0^5 + 2= 2#

#f'(x) = 5x^4#
#f''(x) = 20x^3#

Hence #f''(0) = 0# #-> f(x)# has an infection point at #(0, 2)#

Finally realise , the domain and range of #f(x)# is # (-oo, +oo) #

These attributes of #f(x)# on the real xy-plane can be seen on the graph below.

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