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

Question

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

Lincoln Anderson · Accepted Answer

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

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

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]}

Elijah Allen · Answer

undefined The function $ f(x) = \frac{\sqrt{x} + 1}{\sqrt{x}} $ has no intercepts.

As for asymptotes, there is a vertical asymptote at $ x = 0 $ because the denominator approaches zero as $ x $ approaches zero. There are no horizontal or slant asymptotes.

Cora Alexander · Answer

undefined To find the intercepts and asymptotes of the function $ f(x) = \frac{\sqrt{x+1}}{\sqrt{x}} $, we need to analyze its behavior.

1. **Intercepts**:
   - **y-intercept**: To find the y-intercept, set $ x = 0 $ and evaluate $ f(x) $.
     $ f(0) = \frac{\sqrt{0+1}}{\sqrt{0}} = \frac{\sqrt{1}}{0} = 1 $
     So, the y-intercept is at (0, 1).
   - **x-intercept**: To find the x-intercept, set $ f(x) = 0 $ and solve for $ x $.
     $ 0 = \frac{\sqrt{x+1}}{\sqrt{x}} $
     This equation has no real solutions, so there are no x-intercepts.

2. **Asymptotes**:
   - **Vertical asymptote**: Vertical asymptotes occur where the function approaches infinity or negative infinity as $ x $ approaches a certain value. In this case, the function has a vertical asymptote at $ x = 0 $ because the denominator becomes zero and the function approaches infinity as $ x $ approaches 0 from the right.
   - **Horizontal asymptote**: To find horizontal asymptotes, we analyze the behavior of the function as $ x $ approaches positive or negative infinity.
     $ \lim_{{x \to \infty}} f(x) = \lim_{{x \to \infty}} \frac{\sqrt{x+1}}{\sqrt{x}} = \lim_{{x \to \infty}} \frac{\sqrt{x} \sqrt{1+\frac{1}{x}}}{\sqrt{x}} = \lim_{{x \to \infty}} \sqrt{1 + \frac{1}{x}} = \sqrt{1 + 0} = 1 $
     $ \lim_{{x \to -\infty}} f(x) = \lim_{{x \to -\infty}} \frac{\sqrt{x+1}}{\sqrt{x}} = \lim_{{x \to -\infty}} \frac{\sqrt{x} \sqrt{1+\frac{1}{x}}}{\sqrt{x}} = \lim_{{x \to -\infty}} \sqrt{1 + \frac{1}{x}} = \sqrt{1 + 0} = 1 $
     Therefore, the function has a horizontal asymptote at $ y = 1 $.