Answer
(a) Smallest possible value for this expression is $2$.
(b)$x+\frac{1}{x}\geq 2$
$x^2 + \frac{x}{x} \geq 2x$
$x^2-2x +1\geq 0$
$x∈(0, +\infty)$
Work Step by Step
(a) Let's put the values provided and consider the value of expression:
$x=1; x+\frac{1}{x} = 1 + \frac{1}{1} = 2$
$x=3; x+\frac{1}{x} = 3 +\frac{1}{3} = 3.(3)$
$x=\frac{1}{2}; x+\frac{1}{x} = \frac{1}{2}+\frac{2}{1} = 2.5$
$x=\frac{9}{10}; x+\frac{1}{x} = \frac{9}{10} + \frac{10}{9} = 2.0(1)$
$x=\frac{99}{100}; x+\frac{1}{x} = \frac{99}{100} + \frac{100}{99} = 2.000(10)$
I will input two more value for a better visualization of output from this expression:
$x=1.001; x+\frac{1}{x} = 1.001+\frac{1}{1.001} = 2.000,000(999,000)$
$x=0.0001; x+\frac{1}{x} = 0.0001+\frac{1}{0.0001} = 10,000.0001$
As you can see the value of this expression varies from $2$ (as $x$ approaches $1$) to $+\infty$ (as $x$ approaches $0$ or $+\infty$). So the minimum value is $2$ when $x=1$.
(b) Let's try to simplify it:
$x+\frac{1}{x}\geq 2$ //Multiply by $x$
$x^2 + \frac{x}{x} \geq 2x$ //Move terms to one side
$x^2-2x +1\geq 0$ //Solve for quadratic inequality
$x=\frac{-b±\sqrt{b^2-4ac}}{2a} = \frac{2+\sqrt{4-4}}{2}=1$
We have: $x∈(-\infty, 0)U(0, +\infty)$, Buy we have restriction $x>0$, so the domain of definition is $x∈(0, +\infty)$
So, for any non-negative $x$ value we get output greater or equal to $2$.
We have also proven it with actual numbers in (a)
