Power functions and the square root function

When $\displaystyle n$ is even ($\displaystyle x^2$, $\displaystyle x^4$, $\displaystyle x^6$, …), the graph of $\displaystyle {y}={x^n}$ looks like the parabola $\displaystyle {y}={x^2}$. On the board – $\displaystyle {y}={x^2}$, and $\displaystyle {y}={x^4}$ (the dashed one):

• The graph passes through $\displaystyle (-1,1)$, $\displaystyle (0,0)$ and $\displaystyle (1,1)$.
• The function is even: $\displaystyle {(-x)^n}={x^n}$ when $\displaystyle n$ is even.
• It decreases where $\displaystyle x<0$ and increases where $\displaystyle x>0$; $\displaystyle (0,0)$ is a minimum point.
• It is never negative: positive at every $\displaystyle x\neq0$.
• The greater $\displaystyle n$ is, the flatter the graph between $\displaystyle -1$ and 1, and the steeper outside them.

When $\displaystyle n$ is odd ($\displaystyle x^3$, $\displaystyle x^5$, …), the graph of $\displaystyle {y}={x^n}$ rises all the time. On the board – $\displaystyle {y}={x^3}$, and $\displaystyle {y}={x^5}$ (the dashed one):

• The graph passes through $\displaystyle (-1,-1)$, $\displaystyle (0,0)$ and $\displaystyle (1,1)$.
• The function is odd: $\displaystyle {(-x)^n}={-x^n}$ when $\displaystyle n$ is odd.
• It increases everywhere, so it has no extremum point. At $\displaystyle (0,0)$ the graph goes flat for a moment and goes on rising.
• Negative where $\displaystyle x<0$, positive where $\displaystyle x>0$.

The function $\displaystyle {f(x)}={\sqrt{x}}$ is defined only where $\displaystyle x\geq0$, because a negative number has no square root. Each of its values is the non-negative number whose square is $\displaystyle x$:

• The domain: $\displaystyle x\geq0$. The graph starts at $\displaystyle (0,0)$.
• The function increases on its whole domain, and more slowly the greater $\displaystyle x$ is: from 0 to 1 it rises by 1, and from 1 to 4 also by just 1.
• Positive where $\displaystyle x>0$.
• Neither even nor odd: at $\displaystyle -x$ it is not defined at all.
• $\displaystyle (0,0)$ is a minimum point at the end of the domain: there are no points of the graph before it, and on all the rest of the domain the function is greater. An extremum can also be at an end of the domain, not only where the graph turns.

$\displaystyle {y}={\sqrt{f(x)}}$ is defined where the expression under the root is not negative: solve the inequality $\displaystyle f(x)\geq0$.

• $\displaystyle {y}={\sqrt{x+2}}$: $\displaystyle x+2\geq0$, that is $\displaystyle x\geq-2$.
• $\displaystyle {y}={\sqrt{4-x}}$: $\displaystyle 4-x\geq0$, that is $\displaystyle x\leq4$.

The solutions of $\displaystyle {x^n}={a}$ are the $\displaystyle x$-coordinates of the meeting points of $\displaystyle {y}={x^n}$ with the line $\displaystyle {y}={a}$. So the number of solutions depends on the shape of the graph:

• $\displaystyle n$ even: when $\displaystyle a>0$ two opposite solutions, when $\displaystyle {a}={0}$ one solution, and when $\displaystyle a<0$ no solution. $\displaystyle {x^4}={16}$ gives $\displaystyle {x}={\pm2}$.
• $\displaystyle n$ odd: one solution for every $\displaystyle a$. $\displaystyle {x^3}={-8}$ gives $\displaystyle {x}={-2}$, because $\displaystyle {(-2)^3}={-8}$.

On the board – four graphs. Their formulas, not in order:
$\displaystyle {y}={x-1}$
$\displaystyle {y}={(x-1)^2}$
$\displaystyle {y}={(x-1)^3}$
$\displaystyle {y}={-(x-1)^4}$
Write on the board each graph’s formula. Use the shape of the graph and its value at $\displaystyle {x}={0}$ and at $\displaystyle {x}={2}$.

$\displaystyle {f(x)}={x^6}$
$\displaystyle {g(x)}={x^7}$
For each of the functions: is it even or odd?
Where does it increase and where does it decrease?
Does it have an extremum point?

On the board – the graphs of $\displaystyle {y}={\sqrt{x}}$, $\displaystyle {y}={\sqrt{x+1}}$ and $\displaystyle {y}={\sqrt{x-1}}$, not in order.
Write on the board each graph’s formula. Use the point where each graph starts.

Find the domain of each function.

$\displaystyle {y}={\sqrt{x-3}}$

$\displaystyle {y}={\sqrt{6-2x}}$

What is the extremum point of $\displaystyle {y}={\sqrt{x-3}}$, and what kind is it?

Solve the equations.