To find the values of x that guarantee that f(x) is within a certain distance of 5, we can use the definition of a limit.

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

|f(x) - 5| < 0.6

Since f(x) = 3x + 2, we can substitute f(x) into the inequality:

|3x + 2 - 5| < 0.6

Simplifying the inequality, we get:

|-3 + 3x| < 0.6

Now we can solve the inequality for x:

-0.6 < -3 + 3x < 0.6

Adding 3 to all parts of the inequality:

2.4 < 3x < 3.6

Dividing all parts of the inequality by 3:

0.8 < x < 1.2

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

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

|f(x) - 5| < c

Substituting f(x) = 3x + 2:

|3x + 2 - 5| < c

Simplifying the inequality:

|-3 + 3x| < c

Now we solve the inequality:

-c < -3 + 3x < c

Adding 3 to all parts of the inequality:

3 - c < 3x < c + 3

Dividing all parts of the inequality by 3:

(3 - c)/3 < x < (c + 3)/3

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

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