The image shows a set of math problems related to linear and affine functions. We need to solve a few problems. Let's focus on the problems on the left side of the image. Problem 3: Calculate the coefficient of the linear function $f$ in each of the following cases: $f(2) = 0.5$ and $f(5) = 10$. Problem 4: Determine the expression of the function $f$ such that $f(7) = -21$. It appears the problem is incomplete, as we don't know if it's a linear or affine function. Let's assume it's a linear function. Problem 5: What is the number that has an image of 24 for the function $f$? Again, the image is incomplete to give context for the function $f$ in this problem. Let's assume that the function $f$ is defined by $f(x) = 3x$.

AlgebraLinear FunctionsAffine FunctionsFunction EvaluationSolving Equations
2025/4/25

1. Problem Description

The image shows a set of math problems related to linear and affine functions. We need to solve a few problems. Let's focus on the problems on the left side of the image.
Problem 3: Calculate the coefficient of the linear function ff in each of the following cases: f(2)=0.5f(2) = 0.5 and f(5)=10f(5) = 10.
Problem 4: Determine the expression of the function ff such that f(7)=21f(7) = -21. It appears the problem is incomplete, as we don't know if it's a linear or affine function. Let's assume it's a linear function.
Problem 5: What is the number that has an image of 24 for the function ff? Again, the image is incomplete to give context for the function ff in this problem. Let's assume that the function ff is defined by f(x)=3xf(x) = 3x.

2. Solution Steps

Problem 3:
A linear function has the form f(x)=axf(x) = ax, where aa is the coefficient.
We have two points on the line: (2,0.5)(2, 0.5) and (5,10)(5, 10).
Using f(2)=0.5f(2) = 0.5, we get 2a=0.52a = 0.5, so a=0.5/2=0.25a = 0.5 / 2 = 0.25.
Using f(5)=10f(5) = 10, we get 5a=105a = 10, so a=10/5=2a = 10 / 5 = 2.
However, if the function is of the form f(x)=ax+bf(x) = ax + b (affine function), then we can write two equations based on the two points:
2a+b=0.52a + b = 0.5
5a+b=105a + b = 10
Subtracting the first equation from the second, we have:
(5a+b)(2a+b)=100.5(5a + b) - (2a + b) = 10 - 0.5
3a=9.53a = 9.5
a=9.5/3=19/6a = 9.5 / 3 = 19/6
Substituting aa back into the first equation:
2(19/6)+b=0.52(19/6) + b = 0.5
19/3+b=1/219/3 + b = 1/2
b=1/219/3=3/638/6=35/6b = 1/2 - 19/3 = 3/6 - 38/6 = -35/6
Thus f(x)=196x356f(x) = \frac{19}{6}x - \frac{35}{6}
Since the problem statement says *linear function*, we use a=0.25a=0.25 and a=2a=2, based on the two values given. There seems to be an issue with this question. But the *coefficient* is aa. If we use the points (2,0.5)(2,0.5) and (5,10)(5,10), then the function is *not* linear, and is an affine function, and as we solved it, the coefficient is 19/619/6.
Problem 4:
If f(x)=axf(x) = ax and f(7)=21f(7) = -21, then 7a=217a = -21, so a=21/7=3a = -21 / 7 = -3. Therefore, f(x)=3xf(x) = -3x.
Problem 5:
If f(x)=3xf(x) = 3x and f(x)=24f(x) = 24, then 3x=243x = 24, so x=24/3=8x = 24 / 3 = 8.

3. Final Answer

Problem 3: If we consider the function to be linear, and based on the isolated values we get from the problem, we can say the coefficient is 0.25 or 2, based on the first or second values. However, if we combine the points (2,0.5) and (5,10), then we can consider that the function is affine instead, and then the coefficient is 19/619/6.
Problem 4: f(x)=3xf(x) = -3x
Problem 5: 8