Question

Evaluate ƒ(x) = 3|x – 2| + 1 for ƒ(–2) and ƒ(1).

Answer 1

f(-2) =13

f(-1) = 4

Explanation:

ƒ(x) = 3|x – 2| + 1

Let x = -2

ƒ(-2) = 3|-2 – 2| + 1

= 3 | -4| +1

Taking the absolute value

= 3*4 +1

= 12 +1 = 13

Let x = 1

ƒ(1) = 3|1 – 2| + 1

= 3 | -1| +1

Taking the absolute value

= 3*1 +1

= 3 +1 = 4

Answer 2

ƒ(x) = 3|x – 2| + 1

To find f(-2) substitute - 2 into f(x)

That's

f(-2) = 3| - 2 - 2 | + 1

= 3| - 4| + 1

But absolute value of any number is positive including negative numbers

That's

| - 4 | = 4

So we have

3(4) + 1

12 + 1

f(-2) = 13

To find f(1) substitute 1 into f(x)

That's

f(1) = 3 | 1 - 2| + 1

= 3 | - 1| + 1

But | - 1| = 1

= 3(1) + 1

= 3 + 1

f(1) = 4

Hope this helps you