Answer:
Critical value f(1)=2.
Minimum at (1,2), function is decreasing for [tex]0<x<1[/tex] and increasing for [tex]x>1.[/tex]
[tex]\left(3,\dfrac{4}{\sqrt{3}}\right)[/tex] is point of inflection.
When 0<x<3, function is concave upwards and when x>3, , function is concave downwards.
Step-by-step explanation:
1. Find the domain of the function f(x):
[tex]\left\{\begin{array}{l}x\ge 0\\x\neq 0\end{array}\right.\Rightarrow x>0.[/tex]
2. Find the derivative f'(x):
[tex]f'(x)=\dfrac{(x+1)'\cdot \sqrt{x}-(x+1)\cdot (\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}=\dfrac{2x-x-1}{2x\sqrt{x}}=\dfrac{x-1}{2x^{\frac{3}{2}}}.[/tex]
This derivative is equal to 0 at x=1 and is not defined at x=0. Since x=0 is not a point from the domain, the crititcal point is only x=1. The critical value is
[tex]f(1)=\dfrac{1+1}{\sqrt{1}}=2.[/tex]
2. For [tex]0<x<1,[/tex] the derivative f'(x)<0, then the function is decreasing. For [tex]x>1,[/tex] the derivative f'(x)>0, then the function is increasing. This means that point x=1 is point of minimum.
3. Find f''(x):
[tex]f''(x)=\dfrac{(x-1)'\cdot 2x^{\frac{3}{2}}-(x-1)\cdot (2x^{\frac{3}{2}})'}{(2x^{\frac{3}{2}})^2}=[/tex]
[tex]=\dfrac{2x^{\frac{3}{2}}-2(x-1)\cdot \frac{3}{2}x^{\frac{1}{2}}}{4x^3}=\dfrac{2x^{\frac{3}{2}}-2\cdot\frac{3}{2}x^{\frac{3}{2}}+ 2\cdot\frac{3}{2}x^{\frac{1}{2}}}{4x^3}=[/tex]
[tex]=\dfrac{-x+3}{4x^{\frac{5}{2}}}.[/tex]
When f''(x)=0, x=3 and [tex]f(3)=\dfrac{3+1}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}.[/tex]
When 0<x<3, f''(x)>0 - function is concave upwards and when x>3, f''(x)>0 - function is concave downwards.
Point [tex]\left(3,\dfrac{4}{\sqrt{3}}\right)[/tex] is point of inflection.
