Functions and generalities: answer key for 3rd grade exercises in PDF.

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Read image and antecedent from the representative curve of a function. Third grade math exercises on the generalities of functions.

Exercise 1:

a. We have h(0)= – 1.

b. The numbers 2 and – 2 have as image 0 by the function f.

c. h(4)=3.5 and h(-3)=1.2 .

Exercise 2:

Exercise 3:

A BOX IS MADE IN A SQUARE CARDBOARD PLATE OF 20 CM SIDE. FOR THIS WE CUT X CM SQUARES AND WE BEND THEM ALONG THE DOTS.

1. WHY X IS BETWEEN O AND 1O .

because the width of the two squares is 2x if x takes the maximum value of 10 the two squares will touch each other beyond that these two squares will not exist.

2. WHAT IS THE HEIGHT OF THE BOX .

The height of the box is x cm.

3. CALCULATE THE AREA A(x) OF THE SQUARE AT THE BOTTOM OF THE BOX IN CM² .

4. CALCULATE THE VOLUME V(x) OF THE BOX IN CM3 .

5. GRAPHICALLY REPRESENT V(x) FOR THE PREVIOUS VALUES .

6. CONJECTURE THE VALUE X FOR WHICH THE VOLUME IS MAXIMUM.

The volume of the box seems to be maximal for x = 3 cm .

Exercise 4:

Say whether the given graphical representations are, yes or no, representations of functions:

Every antecedent for a function has a unique image.

We deduce that the representative curves of functions are the graphs 1, 3, 4 and 5.

Exercise 5:

Romeo is in R, Juliet in J.

Romeo must go to pick a flower on the wall of roses [AB] and bring it to Juliette, as quickly as possible, thus by the shortest way.

BR = 5 m , AJ = 3 m and AB=10 .

Determine the position of the point M so that its footprint is the shortest path.

Let’s express the length covered by Romeo :

Note with $0\leq\,\,x\leq\,\,10$

Let’s apply the Pythagorean theorem to the rectangular triangles JAM and MBR.

and

$JM^2=3^2+x^2\\$ and

and

$JM=\sqrt{x^2+9}$ and $MR=\sqrt{(10-x)^2+25}$

thus

$Longueur=\sqrt{x^2+9}+\sqrt{(10-x)^2+25}$

For the mathematical part, we use Pythagoras, functions, square roots, fractions and then Thales using the point J’ symmetrical of J with respect to A.

(RJ’) $\cap$ (AB) immediately gives the desired position.

Here is the curve when we vary the point M corresponding to the variations of the length

travelled by Romeo.

Exercise 6:

For his dog, Aicko, Mr. Martin wishes to build a rectangular enclosure along his wall.

It has 21 m of fencing.

He wants to use the 21 m of fencing and give maximum space for Aicko. 1) a. What is the length of the pen if the owner chooses a width of 3m? of 7m ?

$L=21-2\times \,3=15\,m$ for a width of 3 m.

$L=21-2\times \,7=7\,m$ for a width of 7 m.

b. How much room does Aicko have to struggle in these two cases?

$A=3\times \,15=45\,\,m^2$ for a width of 3 m.

$A=7\times \,7=49\,\,m^2$ for a width of 7 m. 2) Mr. Martin wants his dog to have as much space as possible.

Note x the width of the enclosure.

a. Give a frame for x (what are the minimum and maximum widths?)

First the width is a positive number so $x\geq\,\,0$

moreover the length must be positive

which is equivalent to $21-2x\geq\,\,0$

$21\geq\,2x$

$\frac{21}{2}\geq\,\,x$

$x\leq\,\,10,5\,m$

Conclusion: x must be between 0 and 10.5 meters.

b. Express, as a function of x, the length of the enclosure.

c. Then prove the expression for the area of the enclosure as a function of x, is .

$A=l\times \,L\,=x\times \,(21-2x)=21x-2x^2$

Remark:

Here are the variations of the area of the dog’s zone when we vary the width of the rectangle:

We could show that the maximum area is reached when $x=\frac{-21}{-4}=5,25\,m$

and the maximum area of the enclosure is:

$A_{max}=5,25\times \,(21-2\times \,5,25)=55,125\,m^2$

and the length is

$L=21-2\times \,5,25=10,5\,m$

Exercise 7:

a. ABC is an equilateral triangle

so the height (AH) is also a perpendicular bisector, we deduce that H is the middle of [BC].

In ABH rectangle in H according to the direct part of the Pythagorean theorem :

${\color{DarkRed}\,BH=\sqrt{18,75}}$

Exercise 8:

Exercise 9:

The number 3.5 has two priors.

The number – 2 has two antecedents.

The number 2 has three antecedents.

Exercise 10:

We have represented below :

– the line with equation y = x,

– the representative curve of a function f defined on [1 ; 8].

The questions asked will be solved by graphic reading.

1. Answer the following questions with true or false:

		true or false
1.	1 has as image 0 by the function f	TRUE
2.	0 has as image 1 by the function f	FALSE
3.	7 is an antecedent of 4 by the function f	TRUE
4.	3 is an antecedent of 4 by the function f	TRUE
5.	f(3) = 4	TRUE
6.	f(2) = 5	FALSE
7.	f(3) > f(5)	TRUE
8.	2.5 has three antecedents by the function f	TRUE
9.	0.5 has a single antecedent by the function f	TRUE
10.	The equation f(x) = 3 has at least one solution in the interval [1 ; 8]	TRUE
11.	The equation f(x) = x has at least one solution in the interval [1 ; 8]	TRUE
12.	f is increasing on the interval [1 ; 8]	FALSE
13.	If x belongs to the interval [4; 5], then f(x) > x	FALSE
14.	If a and b belong to the interval [3 ; 5] and if a < b, then f(a) < f(b)	FALSE

2. Solve graphically the inequation: f(x) – f(3) > 0. The solution will be given as an interval.

is equivalent to f(x) > f(3)

is equivalent to f(x) > 4

The solution set is the interval ]7;8]

Exercise 11:

$EM=\frac{2}{3}EF=\frac{2}{3}\times \,5,4=3,6\,cm$

c. Using Thales’ theorem, we get:

$EN=EG\times \,\frac{EM}{EF}=7,2\times \,\frac{3,6}{5,4}=4,8\,cm$

Exercise 12:

There are three varieties of tuna caught in French Polynesia:
. Albacore tuna (variety of white tuna)
. Yellowfin tuna (a variety of bluefin tuna)
. Bigeye tuna (variety of bluefin tuna)

1. Graph 1, on the following page, represents the size of the Albacore tuna as a function of its mass
a. Is the size of the albacore tuna proportional to its mass? Justify.

No, the curve is not a straight line passing through the origin so the size of the tuna is not proportional to its mass.

b. The Moana team caught a 22 kg Albacore tuna.
Determine graphically its size.

Its size is 100 cm.

c. The Teiki team caught a 70 cm albacore tuna. Determine graphically its mass.

Its mass is 7 kg.

2. The mass of yellowfin tuna represents on average 17% of the total mass of the three tuna species
fished.
Graph 2, on the following page, represents the mass of yellowfin tuna caught in relation to the total mass
of tuna caught.
a. Is the mass of yellowfin tuna proportional to the total mass of tuna caught? Justify.

Yes, because the curve is a straight line passing through the origin of the reference frame.

b. Moana’s team caught 400 kg of tuna. Calculate the mass of yellowfin tuna caught.

The mass of yellowfin tuna caught is 69 kg.

Exercise 13:

a) h(8) = 2 .

b) h(-1)=1 .

c) The antecedents of 0 are x = 3 and x = 7 .

d) h(-3) = 4 .

e) The antecedents of – 2 are x = 4 and x = 6 .

f ) The antecedents are x = -2 ; x = 0 ; x = 2 ; x = 8 .

Exercise 14:

1 ) No the speed was not constant since the curve is not a straight line.

2) Yes, the runner stopped between the 20th and 25th minute.

3 ) The image of 5 is 1. This means that in 5 minutes, it has traveled 1 km .

4) The antecedent of 6 is 35, which means that he traveled 6 km in 35 minutes.

5) This climb started at the 10th minute and the length of this side is 2 km.

6) Because the curve is more and more vertical and it covers more distance.

7) During the descent, in 10 minutes, he covered 3 km

Exercise 15:

2) $v(6)=18\pi%5B(1+\frac{6}{6})-1%5D$

$v(6)=18\pi%5B(\frac{6}{6}+\frac{6}{6})-1%5D$

$v(6)=18\pi%5B2-1%5D$

${\color{DarkRed}\,v(6)=18\pi}$

3) $v(6)=18\pi\simeq\,57$

4) The antecedent of 250 is such that $4<\,x<\,5$ .

$v=\frac{d}{t}=\frac{3000}{10}$

$v=300\times \,60\,m/h$

$v=18000\,m/h$

${\color{DarkRed}\,v=18\,km/h}$

The answer key to the math exercises on the generalities of functions in 3rd grade.

After having consulted the answers to these exercises on the generalities of functions in 3rd grade, you can return to the exercises in 3rd grade.

The exercises in the third grade.

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