The second derivative and the type of an extremum

Calculus, grade 11 (5 units) Lesson 5 of 26Learning

The second derivative and the type of an extremum

The type of an extremum from

\displaystyle f''

At a point $\displaystyle x_0$ where $\displaystyle {f'(x_0)}={0}$ , the sign of the second derivative gives the type of the extremum:

\displaystyle f'(x_0)

\displaystyle =

\displaystyle 0,\ f''(x_0)>0

– a minimum point.

\displaystyle f'(x_0)

\displaystyle =

\displaystyle 0,\ f''(x_0)<0

– a maximum point.

The reason: if $\displaystyle f''(x_0)>0$ , the derivative $\displaystyle f'$ rises around $\displaystyle x_0$ . It passes through 0, so it is negative left of $\displaystyle x_0$ and positive right of it – the function falls and then rises. In terms of concavity: the graph is concave up, and where the tangent is horizontal the graph is above the tangent – a minimum.

Example:

\displaystyle x^3-3x+1

The example below finds the type of the extremum points of $\displaystyle {f(x)}={x^3-3x+1}$ with $\displaystyle f''$ :

$\displaystyle {f'(x)}={3x^2-3}$ , and $\displaystyle {f'(x)}={0}$ where $\displaystyle {x}={1}$ or $\displaystyle {x}={-1}$ .
$\displaystyle {f''(x)}={6x}$ .
$\displaystyle {f''(1)}={6>0}$ : a minimum at $\displaystyle {x}={1}$ , $\displaystyle {f(1)}={-1}$ .
$\displaystyle {f''(-1)}={-6<0}$ : a maximum at $\displaystyle {x}={-1}$ , $\displaystyle {f(-1)}={3}$ .

There is no need for a table of signs of $\displaystyle f'$ : it is enough to put the points into $\displaystyle f''$ .

\displaystyle f(x)=x^{3}-3x+1

\displaystyle A=(-1,3)

\displaystyle B=(1,-1)

When

\displaystyle {f''(x_0)}={0}

The rule works one way only: it says what follows from the sign of $\displaystyle f''$ , but at a minimum point $\displaystyle f''$ need not be positive. When $\displaystyle {f'(x_0)}={0}$ and also $\displaystyle {f''(x_0)}={0}$ , no conclusion can be drawn about the point. The board below shows three functions, each with $\displaystyle {f'(0)}={0}$ and $\displaystyle {f''(0)}={0}$ :

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

\displaystyle g(x)=-x^{4}

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

$\displaystyle {f(x)}={x^4}$ – a minimum at $\displaystyle (0,0)$ .
$\displaystyle {g(x)}={-x^4}$ – a maximum at $\displaystyle (0,0)$ .
$\displaystyle {h(x)}={x^3}$ – no extremum: a horizontal inflection point.

In such a case the type is found another way – from the table of signs of $\displaystyle f'$ .

Ways to tell the type of an extremum

There are several ways to tell the type of an extremum point; choose the most convenient:

A table of signs of $\displaystyle f'$ : from + to − – a maximum, from − to + – a minimum. Always works.
The sign of $\displaystyle f''$ at the point: handy when $\displaystyle f''$ is easy to work out; gives no answer when $\displaystyle {f''(x_0)}={0}$ .
Values of $\displaystyle f$ : compare $\displaystyle f(x_0)$ with the values of $\displaystyle f$ at nearby points on both sides.
The graph of the derivative: the graph of $\displaystyle f'$ crosses the $\displaystyle x$ -axis upwards – a minimum, downwards – a maximum.

Exercise 1

The function is $\displaystyle {f(x)}={2x^3-9x^2+12x}$ .
Its derivative is $\displaystyle {f'(x)}={6x^2-18x+12}$ .

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

$\displaystyle {f''(x)}={12x-18}$ .
Work out $\displaystyle f''$ at the two points you found.

$\displaystyle f(x)=2x^{3}-9x^{2}+12x$

/ – 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

$\displaystyle f(x)=2x^{3}-9x^{2}+12x$

/ – 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 points on the graph.

\displaystyle f(x)=2x^{3}-9x^{2}+12x

Write the type of each extremum point, from the sign of $\displaystyle f''$ .

Exercise 2

The function is $\displaystyle {f(x)}={x+\frac{4}{x}}$ .
Its derivative is $\displaystyle {f'(x)}={1-\frac{4}{x^2}}$ .

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

Find $\displaystyle f''(x)$ .

$\displaystyle f(x)=x+\frac{4}{x}$

/ – 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

Work out $\displaystyle f''$ at the two points you found.

$\displaystyle f(x)=x+\frac{4}{x}$

/ – 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

$\displaystyle f(x)=x+\frac{4}{x}$

/ – 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 and the type of each.
The value at the maximum is smaller than the value at the minimum. How does that fit the graph?

Exercise 3

The function is $\displaystyle {f(x)}={x^5-5x^4}$ .
Its derivative is $\displaystyle {f'(x)}={5x^4-20x^3}={5x^3(x-4)}$ .

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

$\displaystyle {f''(x)}={20x^3-60x^2}$ .
Work out $\displaystyle f''(4)$ and $\displaystyle f''(0)$ .

$\displaystyle f(x)=x^{5}-5x^{4}$

/ – 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

$\displaystyle f(x)=x^{5}-5x^{4}$

/ – 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

What type of extremum is at $\displaystyle {x}={4}$ ?
At $\displaystyle {x}={0}$ the second derivative does not help. Build a table of signs of $\displaystyle f'$ , and decide what is at $\displaystyle {x}={0}$ .

Exercise 4

The family of functions is $\displaystyle {f(x)}={x^3-ax^2+3x}$ , where $\displaystyle a$ is a parameter.
The derivative is $\displaystyle {f'(x)}={3x^2-2ax+3}$ .

The function has two extremum points when the equation $\displaystyle {f'(x)}={0}$ has two solutions, that is, when the discriminant $\displaystyle 4a^2-36$ is positive.
Solve the inequality.

What happens when $\displaystyle {a}={3}$ ? And when $\displaystyle -3<a<3$ ?
In each case write how many extremum points the function has, and whether it has a horizontal inflection.

Formulas: calculus, grade 11 (5 units)

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Rules of differentiation and the tangent line

The derivative of a power
$\displaystyle (x^n)'$ $\displaystyle =$ $\displaystyle nx^{n-1}$
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 quotient
$\displaystyle \left[\frac{f(x)}{g(x)}\right]'$ $\displaystyle =$ $\displaystyle \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$
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 tangent line at the point $\displaystyle (x_1,y_1)$ on the graph
$\displaystyle y-y_1=f'(x_1)(x-x_1)$

The second derivative

The second derivative – the derivative of the derivative
$\displaystyle f''(x)$ $\displaystyle =$ $\displaystyle \left(f'(x)\right)'$
The function is concave up ( $\displaystyle \cup$ )
$\displaystyle f''(x)>0$
The function is concave down ( $\displaystyle \cap$ )
$\displaystyle f''(x)<0$
A possible inflection point – check that the sign of $\displaystyle f''$ changes there
$\displaystyle f''(x_0)=0$
A minimum point
$\displaystyle f'(x_0)=0,\ f''(x_0)>0$
A maximum point
$\displaystyle f'(x_0)=0,\ f''(x_0)<0$
Acceleration – the derivative of the speed
$\displaystyle a(t)=v'(t)=s''(t)$

Asymptotes

A vertical asymptote – $\displaystyle f(x)$ grows or falls without end as $\displaystyle x$ approaches $\displaystyle a$
$\displaystyle x=a$
A horizontal asymptote – $\displaystyle f(x)$ approaches $\displaystyle b$ as $\displaystyle x\to\infty$ or as $\displaystyle x\to-\infty$
$\displaystyle y=b$

Roots

The root of a square
$\displaystyle \sqrt{x^2}$ $\displaystyle =$ $\displaystyle |x|$
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
$\displaystyle (\sqrt{f(x)})'$ $\displaystyle =$ $\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}$ $\displaystyle f(x)>0$

Radians and periods

Half a turn in radians
$\displaystyle \pi$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The length of an arc of $\displaystyle \alpha$ radians in a circle of radius $\displaystyle R$
$\displaystyle l$ $\displaystyle =$ $\displaystyle \alpha R$
The period of $\displaystyle \sin x$ and of $\displaystyle \cos x$ is $\displaystyle 2\pi$
$\displaystyle \sin(x+2\pi)$ $\displaystyle =$ $\displaystyle \sin x$
The period of $\displaystyle \tan x$ is $\displaystyle \pi$
$\displaystyle \tan(x+\pi)$ $\displaystyle =$ $\displaystyle \tan x$
The limit behind the derivative of $\displaystyle \sin x$ ( $\displaystyle x$ in radians)
$\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$

Trigonometric identities

The Pythagorean identity
$\displaystyle \sin^2\alpha+\cos^2\alpha$ $\displaystyle =$ $\displaystyle 1$
Tangent
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
Sine – an odd function
$\displaystyle \sin(-\alpha)$ $\displaystyle =$ $\displaystyle -\sin\alpha$
Cosine – an even function
$\displaystyle \cos(-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
The sine of the supplement to $\displaystyle \pi$
$\displaystyle \sin(\pi-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The cosine of the complement to $\displaystyle \frac{\pi}{2}$
$\displaystyle \cos\left(\frac{\pi}{2}-\alpha\right)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of a double angle
$\displaystyle \sin2\alpha$ $\displaystyle =$ $\displaystyle 2\sin\alpha\cdot\cos\alpha$
The cosine of a double angle
$\displaystyle \cos2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$
The difference of two sines
$\displaystyle \sin\alpha-\sin\beta$ $\displaystyle =$ $\displaystyle 2\cos\frac{\alpha+\beta}{2}\cdot\sin\frac{\alpha-\beta}{2}$
The sine of a sum and of a difference
$\displaystyle \sin(\alpha\pm\beta)$ $\displaystyle =$ $\displaystyle \sin\alpha\cdot\cos\beta\pm\cos\alpha\cdot\sin\beta$

Derivatives of the trigonometric functions

The derivative of sine ( $\displaystyle x$ in radians)
$\displaystyle (\sin x)'$ $\displaystyle =$ $\displaystyle \cos x$
The derivative of cosine
$\displaystyle (\cos x)'$ $\displaystyle =$ $\displaystyle -\sin x$
The derivative of tangent
$\displaystyle (\tan x)'$ $\displaystyle =$ $\displaystyle \frac{1}{\cos^2x}$