For the function f(x) = #sqrtx+1/sqrtx# what are the intercepts and asymptotes ?

1 Answer
Jun 18, 2017

Answer:

#f(x)# has a vertical asymptote #x=0# and no other linear asymptotes.

It does not intercept the #x# or #y# axes.

Explanation:

Given:

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

Note that:

  • #sqrt(x)# only takes real values when #x >= 0#.

  • #1/sqrt(0) = 1/0# is undefined.

So we find that #f(x)# is well defined precisely when #x > 0#.

As #x->0^+#:

  • #sqrt(x)->0#

  • #1/sqrt(x)->oo#

So #f(x)# has a vertical asymptote #x=0#.

For any #x > 0# both #sqrt(x) > 0# and #1/sqrt(x) > 0#, so #f(x)# has no intercepts with the #x#-axis.

As #x->oo#:

  • #sqrt(x)->oo#

  • #1/sqrt(x)->0#

So #f(x)# is asymptotic to the function #y = sqrt(x)#. It has no horizontal or oblique (slant) asymptotes.

By symmetry, the minimum value of #f(x)# occurs where #sqrt(x) = 1/sqrt(x)#, i.e. where #x=1#.

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

We now have enough information to draw the graph of #f(x)#...
graph{(y-sqrt(x)-1/sqrt(x))((x-1)^2+(y-2)^2-0.01) = 0 [-4.13, 15.87, -1.64, 8.36]}