The graph of √f(x) from the graph of f

The graph of $\displaystyle {g(x)}={\sqrt{f(x)}}$ can be sketched from the graph of $\displaystyle f$ alone, since each property of $\displaystyle g$ comes from one of $\displaystyle f$:

• the domain: $\displaystyle f(x)\geq0$ – the parts of the graph of $\displaystyle f$ on or above the $\displaystyle x$-axis;
• the zeros: those of $\displaystyle f$;
• rising and falling: $\displaystyle {g'(x)}={\frac{f'(x)}{2\sqrt{f(x)}}}$ has the sign of $\displaystyle f'(x)$ wherever $\displaystyle f(x)>0$, so $\displaystyle g$ rises and falls where $\displaystyle f$ does, and has its extremum points at the same $\displaystyle x$ (inside its domain);
• vertical tangents: at a zero where the graph of $\displaystyle f$ crosses the axis, $\displaystyle g'$ grows without end – the tangent to $\displaystyle g$ there is vertical;
• above or below $\displaystyle f$: where $\displaystyle f(x)>1$, $\displaystyle \sqrt{f(x)}<f(x)$; where $\displaystyle 0<f(x)<1$, $\displaystyle \sqrt{f(x)}>f(x)$; where $\displaystyle f(x)$ is 0 or 1 the graphs meet;
• asymptotes: where $\displaystyle f(x)$ approaches $\displaystyle b>0$, $\displaystyle g(x)$ approaches $\displaystyle \sqrt{b}$.

The example below sketches $\displaystyle {g(x)}={\sqrt{4+3x-x^2}}$ from the parabola $\displaystyle {f(x)}={4+3x-x^2}$ (dashed):

• $\displaystyle {f(x)}={-(x+1)(x-4)}$ is not negative where $\displaystyle -1\leq x\leq4$: the domain of $\displaystyle g$;
• the zeros $\displaystyle (-1,0)$ and $\displaystyle (4,0)$, with vertical tangents there;
• $\displaystyle f$ has a maximum at $\displaystyle {x}={1.5}$, with $\displaystyle {f(1.5)}={6.25}$, so $\displaystyle g$ has a maximum at $\displaystyle {(1.5,\sqrt{6.25})}={(1.5,2.5)}$;
• the ends $\displaystyle (-1,0)$ and $\displaystyle (4,0)$ are minimum points at the end.

When the extremum point of $\displaystyle f$ is where $\displaystyle f$ is negative, $\displaystyle g$ does not get it. For $\displaystyle {f(x)}={x^2-4}$ and $\displaystyle {g(x)}={\sqrt{x^2-4}}$:

• the domain of $\displaystyle g$: $\displaystyle x\leq-2$ or $\displaystyle x\geq2$;
• the minimum of $\displaystyle f$ at $\displaystyle {x}={0}$ is outside the domain: $\displaystyle g$ has no extremum point inside its domain;
• $\displaystyle (-2,0)$ and $\displaystyle (2,0)$ are minimum points at the end, with vertical tangents.

The board shows $\displaystyle {f(x)}={x^2-2x-3}$.
$\displaystyle {g(x)}={\sqrt{f(x)}}$ is on the board too, hidden: its dot shows it.

Find the domain of $\displaystyle g$.

Mark the zeros of $\displaystyle g$; then show its graph.

$\displaystyle f$ has a minimum point at $\displaystyle {x}={1}$.
Why has $\displaystyle g$ no point there? What are the extremum points of $\displaystyle g$?

The board shows $\displaystyle {f(x)}={-x^2+6x-5}$.
$\displaystyle {g(x)}={\sqrt{f(x)}}$ is hidden on it.

Find the domain of $\displaystyle g$.

Mark the zeros of $\displaystyle g$ and its extremum point inside the domain, from the graph of $\displaystyle f$; then show the graph of $\displaystyle g$.

Write all the extremum points of $\displaystyle g$ and their types.
Where is the graph of $\displaystyle g$ above the graph of $\displaystyle f$?

The board shows $\displaystyle {f(x)}={\frac{4}{x^2+1}}$.
$\displaystyle {g(x)}={\sqrt{f(x)}}$ is hidden on it.

Mark the extremum point of $\displaystyle g$; then show its graph.

The graph of $\displaystyle g$ is above the graph of $\displaystyle f$ where $\displaystyle 0<f(x)<1$.
Solve the inequality $\displaystyle f(x)<1$.

What is the asymptote of $\displaystyle g$? Where do the graphs of $\displaystyle f$ and $\displaystyle g$ meet?