How do you sketch the curve #y=(x+5)^(1/4)# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

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Answer 1 · 2017-06-28T00:03:12

The function #f(x) = (x+5)^(1/4) # is defined and continuous for #(x+5) > 0#, that is for #x in [-5,+oo]#.

At the boundaries of the domain we have:

#lim_(x->5^+) (x+5)^(1/4) = 0#

#lim_(x->+oo) (x+5)^(1/4) = oo#

inside the domain the function has only positive values.

Evaluate the derivatives:

#f'(x) = 1/4(x+5)^(-3/4)#

#f''(x) = -3/8(x+5)^(-7/4)#

So we have that:

#f'(x) > 0# for #x in (-5,+oo)# and #lim_(x->5^+) f'(x) = oo#

#f''(x) < 0# for #x in (-5,+oo)#

We can then conclude that #f(x)# is strictly increasing and has no relative maxima or minima and is concave down in its domain:

graph{(x+5)^(1/4) [-10, 10, -5, 5]}