Evaluating Functions

This lesson is about evaluating functions.

1. Lesson 1. 2 Evaluating Functions   Learning Competency Evaluates a function (M11GM-Ia-2).

2. Evaluating a function means finding the value of the function by replacing or substituting its variable with a given number or expression. For example, the function f(x)=10 + 6x 2 can be evaluated by squaring the input value, multiplying by 6, and then adding the product from 10. In this lesson, you are expected to evaluate function while using the concept of PEMDAS rule (Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction), rules for integers, and law of substitution.

3. Steps on how to evaluate a function   Substitute or replace the input variable in the formula with the value provided. Simplify the expression on the right-hand side of the equation to obtain the result.

4. Example 1: Given g(x) = 5x – 23, find f(4).   Solution: g(4) = 5(4) – 23 Substitute 4 for x in the function. g(4) = 20 – 23 Simplify the expression on the right side g(4) = -3 of the equation by using PEMDAS and rules for integers. Note: 5x means “5 times x”.

5. Example 2: Given f(x)=10 + 6 ⌈ x ⌉ 2 — ⌊ x ⌋ , evaluate f at x = -3.   Solution: f(x)=10 + 6 ⌈ x ⌉ 2 ⌊ x ⌋ Substitute -3 for x in the function. f(x)=10 + 6 ⌈ -3 ⌉ 2 ⌊ -3 ⌋ Simplify the expression on the right f(x)=10 + 6 ( -2 ) 2 ( -4 ) side of the equation by using f(x)=10 + 6 (4) — ( -4 ) PEMDAS and rules for integers. f(x)=10 + 24 — ( -4 ) f(x)=38 Notes: Floor function, ⌊ x ⌋ , is the greatest integer that is less than or equal to x . Ceiling function, ⌈ x ⌉ , is the least integer that is greater than or equal to x.  

6. Example 3: Given p(a) = , evaluate the function at a = 2 Solution: p(2) = Substitute 4 for x in the function. p(2) = Simplify the expression on the right side p(2) = of the equation by using PEMDAS and p(2) = rules for integers.  

7. Unlock the Process Fill in the missing number/variable represented by (?) in each given function to unlock the process of evaluating function.  1 . Evaluate f(x) = -5 + 3x for x = 4. 2. Given h(x) = x 2 – 3x + 5, find h(-2)   f(?) = -5 + 3(4) h(-2) = (?) 2 – 3(?) + 5 f(?) = -5 + (?) h(-2) = (?) – (?) + 5 f(?) = (?) h(?) = 15   3. If t(x) = , find t(3). 4. Evaluate g(x) = 2x 3 + x – 7 for x =1   t(3) = g(1) = 2(1) + (?) – 7 t(?) = (?)(1) = (?) + (?) – 7 t(?) = g(1) = (?)