Answer to Question #273986 in Calculus for Josa Kyme Palma

Question

Find the points of inflection and discuss the concavity of the graph of the function

f(x)= (x+1)/√x

Answer 1

Concavity and flection points are investigated using the second derivative and its sign.$f(x)= \frac{x+1}{ \sqrt{x}}={x+1\over x^{1\over 2}}$

Apply quotient rule to get the first derivative :

Let $u=(x+1) \implies u'=1$ and $v=x^{1\over 2} \implies v'={1\over 2}x^{-{1\over 2}}$

$f'(x)={u'v-uv'\over v^2}={1.x^{1\over 2}-(x+1).({1\over 2}x^{-{1\over 2}})\over (x^{1\over 2})^2}={x-1\over 2x^{3\over 2}}$

$f''(x)=-{x-3\over 4x^{5\over 2}}=0 \implies x=3$

If $x$ is to the left of $3$ (say at $x=2$ ), $f''(x)\gt 0$ ,so the graph of $f(x)$ is concave up to the left of 3.

If $x$ is to the right of 3, $f''(x)\lt0$ ,so the graph of $f(x)$ is concave down to the right of 3.

The concavity changes as we cross $x=3$ if there is a point on a graph at $x=3$ ( If 3 is in the domain of $f(x)$ ) then that point is an inflection point

$f(3)={(3+1)\over \sqrt 3}=2.3$

So the point $(3,2.3)$ is an infection point.