Investigating a square root function

Algebra and calculus, grade 10 (4 units) Lesson 18 of 20Learning

Investigating a square root function

The steps of investigating a root function

A square root function is investigated in the same steps as a polynomial, but the domain comes first, and everything else is looked at only inside it.

The domain: what is under the root is greater than 0 or equal to it.
The points on the axes: on the $\displaystyle y$ -axis – only if 0 is in the domain; on the $\displaystyle x$ -axis – solve $\displaystyle {f(x)}={0}$ .
The derivative: bring it to a common denominator, and solve $\displaystyle {f'(x)}={0}$ by the numerator. A solution outside the domain is no extremum point.
Where it increases and decreases: the sign of $\displaystyle f'(x)$ on each interval, in a table.
Extremum points: inside the domain – by the change of sign of $\displaystyle f'(x)$ ; at an end of the domain – by the direction of the function next to the end.
A sketch: start at the end of the domain, and pass through the extremum points and the points on the axes.

Investigating

\displaystyle {f(x)}={\sqrt{x^2-4x+3}}

The example below investigates the function $\displaystyle {f(x)}={\sqrt{x^2-4x+3}}$ .

The domain: $\displaystyle x^2-4x+3\geq0$ , that is $\displaystyle x\leq1$ or $\displaystyle x\geq3$ .
On the $\displaystyle y$ -axis: $\displaystyle {f(0)}={\sqrt{3}}$ , the point $\displaystyle (0,\sqrt{3})$ .
On the $\displaystyle x$ -axis: $\displaystyle {x^2-4x+3}={0}$ , the points $\displaystyle (1,0)$ and $\displaystyle (3,0)$ – the two ends of the domain.
The derivative: $\displaystyle {f'(x)}={\frac{2x-4}{2\sqrt{x^2-4x+3}}}={\frac{x-2}{\sqrt{x^2-4x+3}}}$ .
$\displaystyle {f'(x)}={0}$ only when $\displaystyle {x}={2}$ , and 2 is not in the domain: there is no extremum point inside the domain.

	$\displaystyle x<1$	$\displaystyle x>3$
$\displaystyle f'(x)$	−	+
$\displaystyle f(x)$	decreasing	increasing

\displaystyle f(x)=\sqrt{x^{2}-4x+3}

\displaystyle y=x-2

\displaystyle A=(1,0)

\displaystyle B=(3,0)

To the left of 1 the function decreases down to $\displaystyle (1,0)$ , and to the right of 3 it increases from $\displaystyle (3,0)$ : both points are minimum points at an end of the domain.
The values of the function are never negative, so these are also the absolute minimum points.
Does the line $\displaystyle {y}={x-2}$ (the dashed one) meet the graph? Squaring $\displaystyle {\sqrt{x^2-4x+3}}={x-2}$ gives $\displaystyle {x^2-4x+3}={x^2-4x+4}$ , that is $\displaystyle {3}={4}$ . There is no solution: to the right of 3 the graph comes closer and closer to the line, but never meets it.

An extremum at the end of the domain and inside it

The example below investigates the function $\displaystyle {f(x)}={x-2\sqrt{x}}$ .

The domain: $\displaystyle x\geq0$ .
On the axes: $\displaystyle {x}={2\sqrt{x}}$ ; squaring, $\displaystyle {x^2}={4x}$ , so $\displaystyle {x}={0}$ or $\displaystyle {x}={4}$ – the points $\displaystyle (0,0)$ and $\displaystyle (4,0)$ .
The derivative: $\displaystyle {f'(x)}={1-\frac{1}{\sqrt{x}}}$ , and $\displaystyle {f'(x)}={0}$ when $\displaystyle {\sqrt{x}}={1}$ , that is $\displaystyle {x}={1}$ .

	$\displaystyle 0<x<1$	$\displaystyle {x}={1}$	$\displaystyle x>1$
$\displaystyle f'(x)$	−	0	+
$\displaystyle f(x)$	decreasing	$\displaystyle -1$	increasing

\displaystyle f(x)=x-2\sqrt{x}

\displaystyle A=(0,0)

\displaystyle B=(1,-1)

\displaystyle C=(4,0)

$\displaystyle (1,-1)$ is a minimum point, and the absolute minimum too.
$\displaystyle (0,0)$ is a maximum point at the end of the domain: to its right the function decreases, so the values beside it are less than 0.
It is a local maximum only: from $\displaystyle {x}={1}$ the function increases without a bound, and $\displaystyle {f(9)}={3}$ is already greater than 0.

The kind of an extremum at the end of the domain

At an end of the domain the kind of the extremum is decided by the direction of the function next to the end:

A left end, and the function increases from it – a minimum; decreases from it – a maximum.
A right end, and the function increases up to it – a maximum; decreases down to it – a minimum.
An extremum at an end is local: to know whether it is absolute, compare it with the other extremum points and with where the graph goes further on.

Exercise 1

The function is $\displaystyle {f(x)}={\sqrt{9-x^2}}$ .

Find the domain.

Is the function even, odd or neither?
Give a reason.

Find $\displaystyle f'(x)$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {f'(x)}={0}$ .

Mark on the graph the extremum point inside the domain.
What are the extremum points at the ends of the domain, and what kind are they?

\displaystyle f(x)=\sqrt{9-x^{2}}

For which values of $\displaystyle k$ has the equation $\displaystyle {f(x)}={k}$ two solutions?
For which values has it one solution?

Exercise 2

The function is $\displaystyle {f(x)}={-x^2\sqrt{x+5}}$ .

Find the domain and the points where the graph meets the axes.

Are there values of $\displaystyle x$ where $\displaystyle f(x)>0$ ?
Give a reason.

Find $\displaystyle f'(x)$ , over a common denominator.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {f'(x)}={0}$ .

Mark on the graph the extremum points inside the domain.
Say what kind each extremum point is, at the end of the domain too.

\displaystyle f(x)=-x^{2}\sqrt{x+5}

How many solutions has the equation $\displaystyle {-x^2\sqrt{x+5}}={-14}$ ?
Give a reason from the investigation.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The function is $\displaystyle {f(x)}={(x-3)\sqrt{x}}$ .

Find $\displaystyle f'(x)$ , over a common denominator.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {f'(x)}={0}$ .

Mark on the graph the extremum point inside the domain.

\displaystyle f(x)=(x-3)\sqrt{x}

Where does the function increase, and where does it decrease?
What is the extremum point at the end of the domain, and what kind is it?

Exercise 4

The function is $\displaystyle {f(x)}={\sqrt{2x+5}}$ , and the line is $\displaystyle {y}={x+1}$ .

Solve the equation.

What is the point where the line meets the graph?
Explain why the other solution of the equation $\displaystyle {2x+5}={(x+1)^2}$ gives no meeting point.

Formulas: algebra and calculus, grade 10 (4 units)

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Algebraic technique

The square of a sum
$\displaystyle (a+b)^2$ $\displaystyle =$ $\displaystyle a^2+2ab+b^2$
The square of a difference
$\displaystyle (a-b)^2$ $\displaystyle =$ $\displaystyle a^2-2ab+b^2$
The difference of squares
$\displaystyle (a+b)(a-b)$ $\displaystyle =$ $\displaystyle a^2-b^2$
Factoring a trinomial – $\displaystyle x_1$ , $\displaystyle x_2$ the roots of $\displaystyle {ax^2+bx+c}={0}$
$\displaystyle ax^2+bx+c$ $\displaystyle =$ $\displaystyle a(x-x_1)(x-x_2)$
The roots of $\displaystyle {ax^2+bx+c}={0}$
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
The product of roots
$\displaystyle \sqrt{a}\cdot\sqrt{b}$ $\displaystyle =$ $\displaystyle \sqrt{ab}$ $\displaystyle a\geq0,\ b\geq0$
The quotient of roots
$\displaystyle \frac{\sqrt{a}}{\sqrt{b}}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{a}{b}}$ $\displaystyle a\geq0,\ b>0$

The properties of a function

An even function – its graph symmetric about the $\displaystyle y$ -axis
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle f(x)$
An odd function – its graph symmetric about the origin
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle -f(x)$
The $\displaystyle x$ -coordinate of the vertex of $\displaystyle {y}={ax^2+bx+c}$
$\displaystyle x=\frac{-b}{2a}$
The absolute value
$\displaystyle |a|=\begin{cases}a & a\geq0\\-a & a<0\end{cases}$

Transforming a graph

A shift of $\displaystyle k$ units up ( $\displaystyle k<0$ – down)
$\displaystyle y=f(x)+k$
A shift of $\displaystyle p$ units to the right ( $\displaystyle p<0$ – to the left)
$\displaystyle y=f(x-p)$
A reflection in the $\displaystyle x$ -axis
$\displaystyle y=-f(x)$
A reflection in the $\displaystyle y$ -axis
$\displaystyle y=f(-x)$
A stretch ( $\displaystyle |a|>1$ ) or a shrink ( $\displaystyle |a|<1$ )
$\displaystyle y=a\cdot f(x)$

The derivative and the rules of differentiation

The average rate of change from $\displaystyle x_1$ to $\displaystyle x_2$ – the slope of the chord
$\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}$
The derivative at a point – the slope of the tangent
$\displaystyle f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$
The derivative of a power
$\displaystyle (x^n)'$ $\displaystyle =$ $\displaystyle nx^{n-1}$
The derivative of a constant
$\displaystyle (c)'$ $\displaystyle =$ $\displaystyle 0$
The derivative of a constant multiple
$\displaystyle (c\cdot f(x))'$ $\displaystyle =$ $\displaystyle c\cdot f'(x)$
The derivative of a sum and of a difference
$\displaystyle (f(x)\pm g(x))'$ $\displaystyle =$ $\displaystyle f'(x)\pm g'(x)$
The derivative of a product
$\displaystyle (f(x)\cdot g(x))'$ $\displaystyle =$ $\displaystyle f'(x)\cdot g(x)+f(x)\cdot g'(x)$
The derivative of a composite function
$\displaystyle (f(g(x)))'$ $\displaystyle =$ $\displaystyle f'(g(x))\cdot g'(x)$

The tangent to a graph

The slope of the tangent at the point $\displaystyle (x_1,y_1)$ on the graph
$\displaystyle m=f'(x_1)$
The equation of the tangent
$\displaystyle y-y_1=m(x-x_1)$

The derivative of a square root function

The derivative of $\displaystyle \sqrt{x}$
$\displaystyle (\sqrt{x})'$ $\displaystyle =$ $\displaystyle \frac{1}{2\sqrt{x}}$ $\displaystyle x>0$
The derivative of the root of a function
$\displaystyle (\sqrt{f(x)})'$ $\displaystyle =$ $\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}$ $\displaystyle f(x)>0$

Rising, falling and extremum points

The function rises
$\displaystyle f'(x)>0$
The function falls
$\displaystyle f'(x)<0$
A point that may be an extremum
$\displaystyle f'(x)=0$