a. To find the values of x that guarantee f(x) is within 0.6 units of 5, we can set up the inequality:

|f(x) - 5| ≤ 0.6

Substituting f(x) = 3x + 2:

|3x + 2 - 5| ≤ 0.6

Simplifying:

|3x - 3| ≤ 0.6

Now, we can solve this inequality. Splitting it into two cases:

Case 1: 3x - 3 ≤ 0.6

Adding 3 to both sides:

3x ≤ 3.6

Dividing both sides by 3:

x ≤ 1.2

Case 2: -(3x - 3) ≤ 0.6

Expanding the negative sign:

-3x + 3 ≤ 0.6

Subtracting 3 from both sides:

-3x ≤ -2.4

Dividing both sides by -3 (and remember that dividing by a negative number flips the inequality sign):

x ≥ 0.8

Combining the solutions from both cases:

0.8 ≤ x ≤ 1.2

Therefore, the values of x that guarantee f(x) is within 0.6 units of 5 are between 0.8 and 1.2 (inclusive).

b. To find the values of x that guarantee f(x) is within c units of 5, we can set up the inequality:

|f(x) - 5| ≤ c

Substituting f(x) = 3x + 2:

|3x + 2 - 5| ≤ c

Simplifying:

|3x - 3| ≤ c

Now, we can solve this inequality. Splitting it into two cases:

Case 1: 3x - 3 ≤ c

Adding 3 to both sides:

3x ≤ c + 3

Dividing both sides by 3:

x ≤ (c + 3)/3

Case 2: -(3x - 3) ≤ c

Expanding the negative sign:

-3x + 3 ≤ c

Subtracting 3 from both sides:

-3x ≤ c - 3

Dividing both sides by -3 (and remember that dividing by a negative number flips the inequality sign):

x ≥ (3 - c)/3

Combining the solutions from both cases:

(c + 3)/3 ≤ x ≤ (3 - c)/3

Therefore, the values of x that guarantee f(x) is within c units of 5 are between (c + 3)/3 and (3 - c)/3 (inclusive).

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