-4 is the y-value or the function value. Function: Evaluating Example 1: Given f (x) = 5 – x2, find f (- 3). The notation f (- 3) means to replace the independent variable, x with - 3 and calculate the value of the dependent (or function) variable, y. So here, f (- 3) = 5 – (- 3)2 = 5 - 9 = - 4 -4 is the y-value or the function value.

Function: Evaluating Example 2: Given f (x) = (x – 4)2, find f (a). Here, replace x with a: f (a) = (a – 4)2 = a2 – 8a + 16. In this case a numerical y- or functional-value has not been found. However, a formula for it in terms of a has been found.

Here f (a + h) means to replace x with a + h so Function: Evaluating Example 3: Given f (x) = x2 + 2x, find f (a + h) – f (a). Here f (a + h) means to replace x with a + h so f (a + h) = (a + h)2 + 2(a + h) f (a + h) = a2 + 2ah + h2 + 2a + 2h f (a) means to replace x with a so f (a) = a2 + 2a

Function: Evaluating Example 3: Given f (x) = x2 + 2x, find f (a + h) – f (a). To find f (a + h) – f (a), subtract the resulting expressions to get (a2 + 2ah + h2 + 2a + 2h) – (a2 + 2a) = a2 + 2ah + h2 + 2a + 2h – a2 – 2a = 2ah + h2 + 2h So f (a + h) – f (a) = 2ah + h2 + 2h. Here, the difference between the y-value when x = a + h and the y-value when x = a has been found.

Function: Evaluating Example 4: Given find f (2). This function is defined in two pieces. One piece is f (x) = x2. It is defined only for x-values that are  3. The other piece is f (x) = x + 6. It is defined only for x-values that are > 3. Since the function is being evaluated for x = 2 and 2  3, the top piece f (x) = x2 is used. Therefore, f (2) = (2)2 = 4.

Function: Evaluating END OF PRESENTATION