Exercises

Work it out first, then reveal. If you get stuck, the link under the exercise takes you to where it is explained.

Differential calculus

Determine the polynomial function of degree four that is axis-symmetric about the y-axis, goes through the origin and has the minimum point

T (2∣ - 16)

Solution

a = 1

c = - 8

e = 0

, so

f (x) = x^{4} - 8 x^{2}

Determine the polynomial function of degree three with the maximum point

H (0∣4)

and the minimum point

T (2∣0)

Solution

a = 1

b = - 3

c = 0

d = 4

, so

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

Determine the polynomial function of degree three that has the saddle point

S (1∣3)

and goes through the point

P (3∣11)

Solution

a = 1

b = - 3

c = 3

d = 2

, so

f (x) = x^{3} - 3 x^{2} + 3 x + 2

p_{N} (x) = 16 - 2 x

, wanted

p_{N} (3)

Solution

16 - 2 \cdot 3 = 10

p_{N} (x) = 40 - 5 x

Solution

maximum price

40

MU, saturation quantity

x = \frac{40}{5} = 8

p_{N} (x) = 25 - 4 x

, wanted

p_{N}^{'} (x)

and its meaning.

Solution

- 4 \cdot 1 x^{0} = - 4

, 4 MU less per QU

P_{1} (1 ∣ 18)

P_{2} (4 ∣ 12)

Solution

m = \frac{12 - 18}{4 - 1} = - 2

18 = - 2 \cdot 1 + b \Rightarrow b = 20

, so

p_{N} (x) = - 2 x + 20

p_{N} (x) = 12 - 3 x

Solution

0 \leq x \leq 4

p_{A} (x) = 5 + 2 x

, wanted the minimum supply price and

p_{A} (4)

Solution

5

MU and

5 + 2 \cdot 4 = 13

p_{N} (x) = 18 - 2 x

and

p_{A} (x) = 6 + x

Solution

18 - 2 x = 6 + x

x = \frac{12}{3} = 4

p = 6 + 4 = 10

, so

S (4 ∣ 10)