Inequalities: from a graph and by intervals

Calculus, grade 10 (5 units) Lesson 4 of 26Learning

Inequalities: from a graph and by intervals

Solving from a graph

The inequality $\displaystyle f(x)>g(x)$ holds exactly where the graph of $\displaystyle f$ is above the graph of $\displaystyle g$ . The ends of the intervals are the points where the graphs meet.

The board below shows $\displaystyle {f(x)}={x^3}$ and $\displaystyle {g(x)}={x}$ . They meet where $\displaystyle {x^3}={x}$ : at $\displaystyle {x}={-1}$ , $\displaystyle {x}={0}$ , $\displaystyle {x}={1}$ .

\displaystyle f(x)=x^{3}

\displaystyle g(x)=x

The graph of $\displaystyle x^3$ is above the line where $\displaystyle -1<x<0$ and where $\displaystyle x>1$ : these are the solutions of $\displaystyle x^3>x$ .
Everywhere else, apart from the meeting points, it is below it: $\displaystyle x^3<x$ where $\displaystyle x<-1$ and where $\displaystyle 0<x<1$ .

The interval method

An inequality of a polynomial written as a product of factors is solved by the interval method:

Bring everything to one side, so that the other side is 0, and factor.
Mark the zeros on the number line. They divide the line into intervals.
In each interval the sign does not change: substitute one number from the interval, or count how many factors are negative there.
Choose the intervals with the sign asked for. With $\displaystyle \leq$ and $\displaystyle \geq$ the zeros are solutions too.

The example below solves $\displaystyle (x+2)(x-1)(x-4)\leq0$ :

$\displaystyle x$	$\displaystyle x<-2$	$\displaystyle -2<x<1$	$\displaystyle 1<x<4$	$\displaystyle x>4$
the sign	−	+	−	+

The solution: $\displaystyle x\leq-2$ or $\displaystyle 1\leq x\leq4$ .

A squared factor

A factor of even multiplicity does not change sign, so the sign is the same on both sides of its zero.

The example below solves $\displaystyle (x-3)^2(x+1)>0$ :

$\displaystyle (x-3)^2\geq0$ , so the sign is that of $\displaystyle x+1$ , apart from $\displaystyle {x}={3}$ , where the product is 0.
$\displaystyle x+1>0$ where $\displaystyle x>-1$ .
The solution: $\displaystyle -1<x<3$ or $\displaystyle x>3$ . The point $\displaystyle {x}={3}$ is not a solution: there the product is 0, not greater than 0.

The biquadratic equation

A biquadratic equation $\displaystyle {ax^4+bx^2+c}={0}$ has only the powers $\displaystyle x^4$ and $\displaystyle x^2$ . Substituting $\displaystyle {t}={x^2}$ gives a quadratic equation in $\displaystyle t$ .

The example below solves $\displaystyle {x^4-5x^2+4}={0}$ :

$\displaystyle {t}={x^2}$ : $\displaystyle {t^2-5t+4}={0}$ , that is $\displaystyle {(t-1)(t-4)}={0}$ , so $\displaystyle {t}={1}$ or $\displaystyle {t}={4}$ .
$\displaystyle {x^2}={1}$ : $\displaystyle {x}={\pm1}$ . $\displaystyle {x^2}={4}$ : $\displaystyle {x}={\pm2}$ .
A negative value of $\displaystyle t$ would give no $\displaystyle x$ , since $\displaystyle x^2\geq0$ .

The same factoring solves an inequality: $\displaystyle {x^4-5x^2+4}={(x^2-1)(x^2-4)}={(x-1)(x+1)(x-2)(x+2)}$ , and by the interval method $\displaystyle x^4-5x^2+4<0$ where $\displaystyle -2<x<-1$ or $\displaystyle 1<x<2$ .

A rational inequality

The sign of a quotient is like the sign of a product: positive when the numerator and the denominator have the same sign. So an inequality $\displaystyle \frac{f(x)}{g(x)}\geq0$ is solved by the interval method too, with two differences:

The zeros of the denominator are marked on the line as well.
A zero of the denominator is never a solution: the quotient is not defined there.

The example below solves $\displaystyle \frac{2x}{x-1}\leq1$ :

We do not multiply by $\displaystyle x-1$ , since its sign is not known. We bring everything to one side: $\displaystyle \frac{2x}{x-1}-1\leq0$ .
A common denominator: $\displaystyle {\frac{2x-(x-1)}{x-1}}={\frac{x+1}{x-1}\leq0}$ .
The zero of the numerator is $\displaystyle {x}={-1}$ , and of the denominator $\displaystyle {x}={1}$ .
The quotient is negative where $\displaystyle -1<x<1$ . At $\displaystyle {x}={-1}$ it is 0, and at $\displaystyle {x}={1}$ it is not defined.
The solution: $\displaystyle -1\leq x<1$ .

Exercise 1

The functions are $\displaystyle {f(x)}={x^2-4}$ and $\displaystyle {g(x)}={2x-1}$ .

Mark on the board the points where the two graphs meet.

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

\displaystyle g(x)=2x-1

Solve the inequality $\displaystyle f(x)<g(x)$ .
Check on the board: where is the parabola below the line?

Exercise 2

Solve the inequalities by the interval method.

Exercise 3

The function is $\displaystyle {f(x)}={x^4-10x^2+9}$ .

Find the zeros of $\displaystyle f$ .

Solve the inequality $\displaystyle f(x)>0$ .

Exercise 4

Solve the inequalities.

Exercise 5

Solve the inequality.
Careful: multiplying by $\displaystyle x-2$ would reverse the inequality where $\displaystyle x<2$ .

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
$\displaystyle y=b$