Given the functions f: x - 2x + 3 and g: x - x − 1. (a) Find f(−2) (b) Find g⁻¹(x) (c) Find fg(x) (d) Find…

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Given the functions f: x -> 2x + 3 and g: x -> x − 1. (a) Find f(−2) (b) Find g⁻¹(x) (c) Find fg(x) (d) Find the value of x, for which f(x) = g(x). Using only a pencil, ruler and a pair of compasses,

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Question No. 3

(i) Given the functions $f: x \mapsto 2x + 3$ and $g: x \mapsto x - 1$ .
So, $f(x) = 2x + 3$ and $g(x) = x - 1$ .

(a) Find $f(-2)$

Substitute $x = -2$ into the function $f(x)$ .
$f(-2) = 2(-2) + 3$ $f(-2) = -4 + 3$ $f(-2) = -1$
The value of $f(-2)$ is $\boxed{-1}$ .

(b) Find $g^{-1}(x)$
Let $y = g(x)$ , so $y = x - 1$ .
Swap $x$ and $y$ to find the inverse relation.
$x = y - 1$
Solve for $y$ to express $g^{-1}(x)$ .
$y = x + 1$
The inverse function $g^{-1}(x)$ is $\boxed{x + 1}$ .

(c) Find $fg(x)$
Substitute $g(x)$ into $f(x)$ .
$fg(x) = f(g(x)) = f(x - 1)$
Replace $x$ in the expression for $f(x)$ with $(x - 1)$ .
$fg(x) = 2(x - 1) + 3$ $fg(x) = 2x - 2 + 3$ $fg(x) = 2x + 1$
The composite function $fg(x)$ is $\boxed{2x + 1}$ .

(d) Find the value of $x$ , for which $f(x) = g(x)$
Set the expressions for $f(x)$ and $g(x)$ equal to each other.
$2x + 3 = x - 1$
Solve the equation for $x$ .
$2x - x = -1 - 3$ $x = -4$
The value of $x$ is $\boxed{-4}$ .

(ii) Using only a pencil, ruler and a pair of compasses,

(a) Draw the line PQ = 6cm
Draw a straight line segment and mark two points P and Q on it such that the distance between them is 6 cm.

(b) Construct the perpendicular bisector of PQ to meet PQ at M
1. Place the compass needle at P and open it to a radius greater than half the length of PQ. Draw arcs above and below the line segment PQ.
2. Without changing the compass radius, place the needle at Q and draw arcs above and below PQ, intersecting the first set of arcs.
3. Draw a straight line connecting the two points where the arcs intersect. This line is the perpendicular bisector of PQ.
4. Mark the point where the perpendicular bisector intersects PQ as M.
(c) Draw the line PS = 5cm, where S is a point on the perpendicular bisector
1. Place the compass needle at P and open it to a radius of 5 cm.
2. Draw an arc that intersects the perpendicular bisector (constructed in part b).
3. Mark the point of intersection on the perpendicular bisector as S.
4. Draw a straight line segment connecting P and S. The length of PS will be 5 cm.
Question No. 4

(i) Given the function $f(x) = x^2 - 3x - 4$ , for $-2 \le x \le 5$ :

(a) Copy and complete the following table
Calculate $f(x)$ for each given value of $x$ using the formula $f(x) = x^2 - 3x - 4$ .
- For $x = -2$ : $f(-2) = (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6$
- For $x = -1$ : $f(-1) = (-1)^2 - 3(-1) - 4 = 1 + 3 - 4 = 0$
- For $x = 0$ : $f(0) = (0)^2 - 3(0) - 4 = 0 - 0 - 4 = -4$
- For $x = 1$ : $f(1) = (1)^2 - 3(1) - 4 = 1 - 3 - 4 = -6$
- For $x = 2$ : $f(2) = (2)^2 - 3(2) - 4 = 4 - 6 - 4 = -6$
- For $x = 3$ : $f(3) = (3)^2 - 3(3) - 4 = 9 - 9 - 4 = -4$
- For $x = 4$ : $f(4) = (4)^2 - 3(4) - 4 = 16 - 12 - 4 = 0$
- For $x = 5$ : $f(5) = (5)^2 - 3(5) - 4 = 25 - 15 - 4 = 6$
Complete the table with the calculated values.
$x$ -2 -1 0 1 2 3 4 5
$y = f(x)$ 6 0 -4 -6 -6 -4 0 6

(b) Find the minimum value of $y = f(x)$ and the corresponding value of $x$
The minimum value of a quadratic function $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$ .
For $f(x) = x^2 - 3x - 4$ , we have $a=1$ and $b=-3$ .
$x = -\frac{-3}{2(1)} = \frac{3}{2} = 1.5$
Substitute $x = 1.5$ into $f(x)$ to find the minimum value.
$f(1.5) = (1.5)^2 - 3(1.5) - 4$ $f(1.5) = 2.25 - 4.5 - 4$ $f(1.5) = -6.25$
The minimum value of $y = f(x)$ is $\boxed{-6.25}$ at $x = \boxed{1.5}$ .

(c) Solve the equation $f(x) = 0$
Set the function $f(x)$ equal to 0.
$x^2 - 3x - 4 = 0$
Factor the quadratic equation.
$(x - 4)(x + 1) = 0$
Solve for $x$ .
$x - 4 = 0 \quad \text{or} \quad x + 1 = 0$ $x = 4 \quad \text{or} \quad x = -1$
The solutions for $f(x) = 0$ are $\boxed{x = -1 \text{ or } x = 4}$ .

(ii) Given a $2 \times 2$ matrix, P defined by $P = \begin{pmatrix} 3 & 2k \\ k & 6 \end{pmatrix}$ , where $k$ is a constant.

(a) Find P when $k = 2$
Substitute $k = 2$ into the matrix P.
$P = \begin{pmatrix} 3 & 2(2) \\ 2 & 6 \end{pmatrix}$ $P = \begin{pmatrix} 3 & 4 \\ 2 & 6 \end{pmatrix}$
The matrix P when $k=2$ is $\boxed{\begin{pmatrix} 3 & 4 \\ 2 & 6 \end{pmatrix}}$ .

(b) Hence or otherwise, find Adj. P, the adjoint of P.
For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , its adjoint is given by $\text{Adj } A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .
Step 18

Given the functions f: x -> 2x + 3 and g: x -> x - 1. (a) Find f(-2) (b) Find g⁻¹(x) (c) Find fg(x) (d) Find the value of x, for which f(x) = g(x). Using only a pencil, ruler and a pair of compasses,

Step-by-step solution

Substitute $x = -2$ into the function $f(x)$ .

Let $y = g(x)$ , so $y = x - 1$ .

Swap $x$ and $y$ to find the inverse relation.

Solve for $y$ to express $g^{-1}(x)$ .

Substitute $g(x)$ into $f(x)$ .

Replace $x$ in the expression for $f(x)$ with $(x - 1)$ .

Set the expressions for $f(x)$ and $g(x)$ equal to each other.

Solve the equation for $x$ .

Calculate $f(x)$ for each given value of $x$ using the formula $f(x) = x^2 - 3x - 4$ .

Complete the table with the calculated values.

The minimum value of a quadratic function $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$ .

Substitute $x = 1.5$ into $f(x)$ to find the minimum value.

Set the function $f(x)$ equal to 0.

Factor the quadratic equation.

Solve for $x$ .

Substitute $k = 2$ into the matrix P.

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , its adjoint is given by $\text{Adj } A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .

Step 18

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More Mathematics Questions

Given the functions f: x -> 2x + 3 and g: x -> x - 1. (a) Find f(-2) (b) Find g⁻¹(x) (c) Find fg(x) (d) Find the value of x, for which f(x) = g(x). Using only a pencil, ruler and a pair of compasses,

Step-by-step solution

Substitute x=−2x = -2x=−2 into the function f(x)f(x)f(x).

Let y=g(x)y = g(x)y=g(x), so y=x−1y = x - 1y=x−1.

Swap xxx and yyy to find the inverse relation.

Solve for yyy to express g−1(x)g^{-1}(x)g−1(x).

Substitute g(x)g(x)g(x) into f(x)f(x)f(x).

Replace xxx in the expression for f(x)f(x)f(x) with (x−1)(x - 1)(x−1).

Set the expressions for f(x)f(x)f(x) and g(x)g(x)g(x) equal to each other.

Solve the equation for xxx.

Calculate f(x)f(x)f(x) for each given value of xxx using the formula f(x)=x2−3x−4f(x) = x^2 - 3x - 4f(x)=x2−3x−4.