How do you graph $y =(x^(1/2) +1)/(x+1)$ ?

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How do you graph $y =(x^(1/2) +1)/(x+1)$ ?

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Answer 1 · 2015-09-18T23:01:27

Analyse this function to find the domain, some points through which it goes, maximum and asymptotic behaviour.

Explanation:

Let $f(x) = (x^(1/2)+1)/(x+1)$

$x^(1/2)$ only takes Real values for $x >= 0$ .

Hence the domain of $f(x)$ is $[0, oo)$

$f(0) = f(1) = 1$

$f'(x) = (-x^(1/2)+x^(-1/2)-2)/(2(x+1)^2)$

Let $t = x^(1/2)$

Then $f'(x) = 0$ when $-t+1/t-2 = 0$

Multiply this equation through by $-t$ to get:

$t^2+2t-1 = 0$

This has roots:

$t = (-2+-sqrt(2^2-(4xx1xx-1)))/(2*1) = -1+-sqrt(2)$

Since $t = sqrt(x) >= 0$ , $sqrt(x) = -1+sqrt(2)$ .

So $x = (sqrt(2)-1)^2 = 3-2sqrt(2) ~~ 0.17157$

When $x = 3 - 2sqrt(2)$ :

$f(x) = (sqrt(x)+1)/(x+1) = sqrt(2)/(4 - 2sqrt(2)) = (sqrt(2)(2+sqrt(2)))/(2(2-sqrt(2))(2+sqrt(2)))$

$=(sqrt(2)+1)/(4-2) = (sqrt(2)+1)/2 ~~ 1.2071$

As $x->oo$ $f(x) -> 0$ since the denominator has higher degree than the numerator, but quite slowly due to the $x^(1/2)$ degree.

Putting this together, we have a curve only defined for $x >= 0$ , that goes through $(0, 1)$ and $(1, 1)$ , is asymptotic to $y = 0$ and has a maximum at $(3-2sqrt(2), (sqrt(2)+1)/2) ~~ (0.17157, 1.2071)$

graph{(x^(1/2)+1)/(x+1) [-0.552, 4.448, -0.78, 1.72]}

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How do you graph $y =(x^(1/2) +1)/(x+1)$ ?

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