The image contains two exercises. Exercise 2 involves a linear function $f$ such that $f(7) = 14$. We need to: 1. Determine the expression for $f(x)$.

1. Problem Description

The image contains two exercises.

Exercise 2 involves a linear function $f$ such that $f(7) = 14$. We need to:

1. Determine the expression for $f(x)$.

2. Find the number $a$ such that $f(a) = 21$.

3. Sketch the graph of the function $f$.

Exercise 3 involves an affine function $g$ such that $g(0) = 3$ and $g(2) = 7$. We need to:

1. Determine the expression for $g(x)$.

2. Calculate $g(1)$ and $g(3)$.

3. Find the number $a$ such that $f(a) = 21$, where $f$ is the function from Exercise

2.

4. Sketch the graph of the function $g$.

2. Solution Steps

Exercise 2:

1. Since $f$ is a linear function, it can be written in the form $f(x) = kx$, where $k$ is a constant. We are given that $f(7) = 14$, so $7k = 14$. Dividing both sides by 7, we get $k = 2$. Therefore, $f(x) = 2x$.

2. We want to find $a$ such that $f(a) = 21$. Since $f(x) = 2x$, we have $2a = 21$. Dividing both sides by 2, we get $a = \frac{21}{2} = 10.5$.

3. The graph of $f(x) = 2x$ is a straight line passing through the origin $(0, 0)$ and the point $(1, 2)$.

Exercise 3:

1. Since $g$ is an affine function, it can be written in the form $g(x) = mx + b$, where $m$ and $b$ are constants. We are given that $g(0) = 3$, so $m(0) + b = 3$, which means $b = 3$. We are also given that $g(2) = 7$, so $2m + b = 7$. Since $b = 3$, we have $2m + 3 = 7$. Subtracting 3 from both sides, we get $2m = 4$. Dividing both sides by 2, we get $m = 2$. Therefore, $g(x) = 2x + 3$.

2. To calculate $g(1)$, we substitute $x = 1$ into $g(x) = 2x + 3$, so $g(1) = 2(1) + 3 = 2 + 3 = 5$.

To calculate $g(3)$, we substitute $x = 3$ into $g(x) = 2x + 3$, so $g(3) = 2(3) + 3 = 6 + 3 = 9$.

3. We want to find $a$ such that $f(a) = 21$. From Exercise 2, $f(x) = 2x$, so we have $2a = 21$, which gives $a = \frac{21}{2} = 10.5$.

4. The graph of $g(x) = 2x + 3$ is a straight line. We can find two points on the line. We know that $g(0) = 3$, so the point $(0, 3)$ is on the line. We also know that $g(2) = 7$, so the point $(2, 7)$ is on the line.

3. Final Answer

Exercise 2:

1. $f(x) = 2x$

2. $a = 10.5$

3. The graph is a line through $(0, 0)$ and $(1, 2)$.

Exercise 3:

1. $g(x) = 2x + 3$

2. $g(1) = 5$, $g(3) = 9$

3. $a = 10.5$

4. The graph is a line through $(0, 3)$ and $(2, 7)$.