Investigating a square root function

Calculus, grade 10 (5 units) Lesson 20 of 26Learning

Investigating a square root function

The steps of an investigation

A root function is investigated like a polynomial, with two differences: the domain is limited, and its ends play a part.

The domain: what is under the root $\displaystyle \geq0$ .
The points where the graph meets the axes.
The derivative, and the solutions of $\displaystyle {f'(x)}={0}$ . The derivative is not defined at an end where the root is 0.
A table of signs of $\displaystyle f'$ , which starts and ends at the ends of the domain.
Extremum points: a point at an end of the domain is an extremum point too – an extremum at the end – by what the function does beside it.
A sketch of the graph.

Example:

\displaystyle \sqrt{4-x^2}

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

The domain: $\displaystyle 4-x^2\geq0$ , that is $\displaystyle -2\leq x\leq2$ .
The zeros: $\displaystyle (-2,0)$ and $\displaystyle (2,0)$ ; and on the $\displaystyle y$ -axis: $\displaystyle (0,2)$ .
$\displaystyle {f'(x)}={\frac{-2x}{2\sqrt{4-x^2}}}={\frac{-x}{\sqrt{4-x^2}}}$ , which is 0 at $\displaystyle {x}={0}$ .

$\displaystyle x$	$\displaystyle -2$	$\displaystyle -2<x<0$	$\displaystyle 0$	$\displaystyle 0<x<2$	$\displaystyle 2$
$\displaystyle f'(x)$		+	0	−
$\displaystyle f(x)$	minimum at the end	rises	maximum	falls	minimum at the end

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

\displaystyle A=(0,2)

\displaystyle B=(-2,0)

\displaystyle C=(2,0)

$\displaystyle (0,2)$ – a maximum point.
$\displaystyle (-2,0)$ and $\displaystyle (2,0)$ – minimum points at the end: from $\displaystyle -2$ the function rises, and towards 2 it falls.
The graph is half a circle: $\displaystyle {y}={\sqrt{4-x^2}}$ is the upper half of the circle $\displaystyle {x^2+y^2}={4}$ .

Example:

\displaystyle x\sqrt{4-x}

The example below investigates $\displaystyle {f(x)}={x\sqrt{4-x}}$ , a product of $\displaystyle x$ and a root.

The domain: $\displaystyle 4-x\geq0$ , that is $\displaystyle x\leq4$ .
The zeros: $\displaystyle {x}={0}$ and $\displaystyle {x}={4}$ .
By the product rule and the derivative of the root:

f'(x)=\sqrt{4-x}+x\cdot\frac{-1}{2\sqrt{4-x}}=\frac{2(4-x)-x}{2\sqrt{4-x}}=\frac{8-3x}{2\sqrt{4-x}}

$\displaystyle {f'(x)}={0}$ where $\displaystyle {8-3x}={0}$ , that is $\displaystyle {x}={\frac{8}{3}}$ . The denominator is positive, so $\displaystyle f'$ has the sign of $\displaystyle 8-3x$ .

$\displaystyle x$	$\displaystyle x<\frac{8}{3}$	$\displaystyle \frac{8}{3}$	$\displaystyle \frac{8}{3}<x<4$	$\displaystyle 4$
$\displaystyle f'(x)$	+	0	−
$\displaystyle f(x)$	rises	maximum	falls	minimum at the end

\displaystyle f(x)=x\sqrt{4-x}

\displaystyle A=(2.67,3.08)

\displaystyle B=(4,0)

$\displaystyle {f\left(\frac{8}{3}\right)}={\frac{8}{3}\sqrt{\frac{4}{3}}}={\frac{16}{3\sqrt{3}}}={\frac{16\sqrt{3}}{9}\approx3.08}$ : a maximum point.
$\displaystyle (4,0)$ – a minimum point at the end.
As $\displaystyle x$ falls without end, so does $\displaystyle f(x)$ , so the function has no absolute minimum.

Exercise 1

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

Explain why $\displaystyle f$ is defined for every $\displaystyle x$ .
Hint: the discriminant of $\displaystyle x^2-2x+5$ .

Find $\displaystyle f'(x)$ .

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

/ – 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 $\displaystyle {f'(x)}={0}$ .

Work out the value of the function at the point you found.

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

/ – 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

Mark the extremum point of $\displaystyle f$ on the graph.

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

Exercise 2

The function is $\displaystyle {f(x)}={(x-3)\sqrt{x}}$ .
Its domain is $\displaystyle x\geq0$ .

Find $\displaystyle f'(x)$ by the product rule.

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

/ – 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 $\displaystyle {f'(x)}={0}$ .

Work out the value of the function at the point you found.

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

/ – 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

Mark on the graph the zeros and the extremum point you found.

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

Write where $\displaystyle f$ rises and where it falls, and all its extremum points – at the end of the domain too.
Say what kind each one is.

Exercise 3

The function is $\displaystyle {f(x)}={\sqrt{6x-x^2}}$ .

Find the domain of $\displaystyle f$ .

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=\sqrt{6x-x^{2}}$

/ – 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

$\displaystyle {f'(x)}={0}$ where $\displaystyle {x}={3}$ .
Work out $\displaystyle f(3)$ .

$\displaystyle f(x)=\sqrt{6x-x^{2}}$

/ – 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

Mark on the graph the extremum point inside the domain.

\displaystyle f(x)=\sqrt{6x-x^{2}}

Write all the extremum points of $\displaystyle f$ , including those at the end, and say what kind they are.

Exercise 4

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

Find $\displaystyle f'(x)$ .

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

/ – 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 inequality $\displaystyle f'(x)<0$ : where does $\displaystyle f$ fall?

Work out the value of the function at the right end of the interval where it falls.

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

/ – 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

Write the extremum points of $\displaystyle f$ , including the one at the end, and say what kind they are.
Does $\displaystyle f$ have an absolute maximum? Explain.

Exercise 5

The function is $\displaystyle {f(x)}={\sqrt{x-1}+\sqrt{5-x}}$ .
Its derivative is $\displaystyle {f'(x)}={\frac{1}{2\sqrt{x-1}}-\frac{1}{2\sqrt{5-x}}}$ .

Find the domain of $\displaystyle f$ .

$\displaystyle {f'(x)}={0}$ where the two fractions are equal.
Solve the equation.

Work out the value of the function at the point you found.

$\displaystyle f(x)=\sqrt{x-1}+\sqrt{5-x}$

/ – 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

Mark on the graph the extremum point inside the domain.
Work out the values of the function at the ends of the domain, and write all the extremum points and what kind they are.

\displaystyle f(x)=\sqrt{x-1}+\sqrt{5-x}

Formulas: calculus, grade 10 (5 units)

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Algebra

The square of a sum and of a difference
$\displaystyle (a\pm b)^2$ $\displaystyle =$ $\displaystyle a^2\pm2ab+b^2$
The difference of two squares
$\displaystyle a^2-b^2$ $\displaystyle =$ $\displaystyle (a-b)(a+b)$
The cube of a sum and of a difference
$\displaystyle (a\pm b)^3$ $\displaystyle =$ $\displaystyle a^3\pm3a^2b+3ab^2\pm b^3$
The sum and the difference of two cubes
$\displaystyle a^3\pm b^3$ $\displaystyle =$ $\displaystyle (a\pm b)(a^2\mp ab+b^2)$
The quadratic formula for $\displaystyle {ax^2+bx+c}={0}$
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
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 properties of a function

An even function – the graph is symmetric about the $\displaystyle y$ -axis
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle f(x)$
An odd function – the graph is symmetric about the origin
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle -f(x)$
The root of a square
$\displaystyle \sqrt{x^2}$ $\displaystyle =$ $\displaystyle |x|$
The absolute value
$\displaystyle |a|=\begin{cases}a & a\geq0\\-a & a<0\end{cases}$

Transformations of a graph

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

The derivative

The average rate of change from $\displaystyle x_1$ to $\displaystyle x_2$ – the slope of the secant
$\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}$
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 instantaneous speed – the derivative of the distance with respect to time
$\displaystyle v(t)$ $\displaystyle =$ $\displaystyle s'(t)$

Rules of differentiation

The derivative of a product of functions
$\displaystyle [f(x)\cdot g(x)]'$ $\displaystyle =$ $\displaystyle f'(x)\cdot g(x)+f(x)\cdot g'(x)$
The derivative of a quotient of functions
$\displaystyle \left[\frac{f(x)}{g(x)}\right]'$ $\displaystyle =$ $\displaystyle \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$
The derivative of a composite function – the outer derivative times the inner derivative
$\displaystyle [f(u(x))]'$ $\displaystyle =$ $\displaystyle f'(u)\cdot u'(x)$
The derivative of $\displaystyle \frac{1}{x}$
$\displaystyle \left(\frac{1}{x}\right)'$ $\displaystyle =$ $\displaystyle -\frac{1}{x^2}$
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
$\displaystyle (\sqrt{f(x)})'$ $\displaystyle =$ $\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}$ $\displaystyle f(x)>0$

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
$\displaystyle y-y_1=m(x-x_1)$

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
$\displaystyle f'(x)=0$

Asymptotes

A vertical asymptote – $\displaystyle a$ a zero of the denominator that is not a zero of the numerator
$\displaystyle x=a$
A horizontal asymptote – $\displaystyle b$ the value the function approaches as $\displaystyle x$ grows or falls without end
$\displaystyle y=b$