Derivative functions

Exercise 1

$f' (x) = 4 x^{3} - 16 x = 0$ when $x = 0 \lor x = \pm 2$ .
Sign plot of $f'$ or graph of $f$ : min. $f (-2) = -16$ , max. $f (0) = 0$ and min. $f (2) = -16$ .

Exercise 2

a

The x-intersects of $f$ are $(\pm 20, 0)$ .
The x-intersects of $g$ are $(\pm 20, 0)$ and $(10, 0)$ .

b

You do not need to differentiate $f$ : the graph is an open down parabola with max. $f (0) = 4000$ .
For $g$ the following is true: $g' (x) = 3 x^{2} - 20 x - 400 = 0$ when $x = \frac{20 \pm \sqrt{5200}}{6}$ .
The extrema of $g$ are therefore: max. $g (-8,69) \approx 6064,60$ and min. $g (15,35) \approx -879,42$ .

c

$x \leq -20 \lor 0 \leq x \leq 20$

Exercise 3

Sports field: length $l$ m and width $2 r$ m where $r$ is the radius of each of the two half circles.
The following must be true: $2 l + 2 π r = 400$ , and therefore $l = 200 - π r$ .
The surface area of the sports field is: $A = l \cdot 2 r = (200 - π r) \cdot 2 r = 400 r - 2 π r^{2}$ .
Calculating te maximum: $A' (r) = 400 - 4 π r = 0$ gives you $r = \frac{100}{π}$ .
The sports field therefore has a length of $100$ m and a width of $\frac{200}{π}$ m.

Exercise 4

a

$\frac{(400 - 200)}{(500 - 400)} = 2$ euro per kg.

b

Use your graphing calculator to make a table of $T C (q)$ and compare the values to those in the table.

c

$T P = 225 q - T C = 225 q - (10 q^{3} - 60 q^{2} + 130 q) = -10 q^{3} + 60 q^{2} + 95 q$

d

$M G (q) = T P' (q) = -30 q^{2} + 120 q + 95$
$M G$ is the speed with which the profit changes as $q$ increases.

e

$M G = 0$ when $q = \frac{-120 \pm \sqrt{25800}}{60}$ .
The maximum profit is about $M G (4,68) \approx 733,71$ .

Exercise 5

a

$f' (x) = 4 x^{3} - 2 a x = 0$ gives you $x = 0 \lor x = \pm \sqrt{(\frac{1}{2} a)}$ .
This means that the function will only have a minimum other than $0$ when $a > 0$ . In that case the minimum is $f (\sqrt{\frac{1}{2} a}) = {(\frac{1}{2} a)}^{2} - a \cdot \frac{1}{2} a = - \frac{1}{4} a^{2}$ .
The minimum is equal to $-1$ if $- \frac{1}{4} a^{2} = -1$ and therefore if $a = 2$ ( $a = -2$ is not allowed).

b

$f' (1) = 4 - 2 a$ and $f (1) = 1 - a$ , so the tangent must have equation $y = (4 - 2 a) x + a - 3$ .
At the same time, $(0, 4)$ must be a solution of this equation: $4 = a - 3$ gives you $a = 7$ .

Exercise 6

a

$f' (x) = 3 x^{2} - 12 p x = 0$ gives you $x = 0 \lor x = 4 p$ .
When $p \neq 0$ the graph of $f$ always has two extrema.

b

$f (0) = -16 \neq -32$
$f (4 p) = -32$ gives you $64 p^{3} - 96 p^{3} - 16 = - 32$ and therefore $32 p^{3} = 16$ and $p = \sqrt[3]{\frac{1}{2}}$ .
This is a minimum.

Solutions to the exercises