GENERAL MATH

PLEASE EXPLAIN HOW'D YOU GET THE ANSWER IN THE SUPPLEMENTARY EXERCISES. HELP ME UNDERSTAND.

lesson:

Lesson 2: Evaluating Functions Learning Outcome(s): At the end of the lesson, the learner is able to evaluate functions and solve problems involving functions. Lesson Outline: 1. Evaluating functions Evaluating a function means replacing the variable in the function, in this case x, with a value from the function's domain and computing for the result. To denote that we are evaluating fat a for some a in the domain of f, we write f(a). Example 1. Evaluate the following functions at x = 1.5: (9) f(x) = 2x + 1 (h) q(x) = x2 - 2x +2 (1) g(x) = Vx + 1 COP ()) r(x) = 2x+1 x-1 (k) F(x) = (x] + 1, where (x] is the greatest integer function. Solution. Substituting 1.5 for x in the functions above, we have (a) f(1.5) = 2(1.5) + 1 = 4 (b) q(1.5) = (1.5)2 - 2(1.5) + 2 = 2.25 - 3 + 2 = 1.25 (c) g(1.5) = V1.5 + 1 = v2.5 (d) r(1.5) = 2(1.5)+1 3+1 (1.5)-1 0.5 (e) F(1.5) = (1.5) +1=1+1 = 2 Example 2. Find g(-4) and r(1) where g and r are as defined in the previous example. Solution. This is not possible because -4 is not in the domain of g(x) and 1 is not in the domain of r(x). Example 3. Evaluate the following functions, where f and q are as defined in Example 1. (a)f (3x - 1) (b) q(2x + 3) Solution. (a) f(3x - 1) = 2(3x - 1) + 1 = 6x - 2+ 1 = 6x - 1 (b) q(2x + 3) = (2x + 3)? - 2(2x +3) + 2 = (4x2 + 12x +9) -4x - 6+2 = 4x + 8x + 5Solved Examples 1. Evaluate the following functions at x=3. (a) f(x) = x -3 (b) g(x) = x2 - 3x +5 (c) h(x) = Vx3+ x+3 (d) p(x) =* +1 x-4 (e) f(x) = |x -5| where |x - 5| means the absolute value of x - 5. Solution. (a) f(3) = 3 -3 = 0 (b) g(3) = (3)2 - 3(3) +5 =9-9+5 =5 (c) h(3) = V(3)3 + 3 +3 = V27+6 = V33 OPY 3-4 (d) p(3) = (3)-+1 _ 10 = -10 (e) f(3) = 13 - 51 = 1-21 = 2 2. For what values of x can we not evaluate the function f(x) = +? Solution. The domain of the function is given by {x : x E R, x # +2). Since 2 and -2 are not in the domain, we cannot evaluate the function at x = -2,2. 3. Evaluate f(a + b) where f(x) = 4x2 - 3x. Solution. f(a + b) = 4(a + b)2 - 3(a+ b) = 4(a2 + 2ab + b2) - 3a -3b = 4a- - 3a + Bab - 3b + 4b 4. Suppose that s(T) is the top speed (in km per hour) of a runner when the temperature is T degrees Celsius. Explain what the statements s(15) = 12 and s(30) = 10 mean. Solution. The first equation means that when the temperature is 15 C, then the top speed of a runner is 12 km per hour. However, when temperature rises to 30 C, the top speed is reduced to 10 km per hour. 5. The velocity V (in m/s) of a ball thrown upward t seconds after the ball was thrown is given by V(t) = 20 - 9.8t. Calculate V(0) and V(1), and explain what these results mean. Solution. V(0) = 20 - 9.8(0) = 20 and V(1) = 20 - 9.8(1) = 10.2. These results indicate that the initial velocity of the ball is 20 m/s. After 1 second, the ball is traveling more slowly, at 10.2 m/s.Lesson 2 Supplementary Exercises 1. Evaluate the following functions at x = -4. (a) f(x) = x3 - 64 (b) g(x) = (x3 - 3x2 + 3x - 1| (c) r(x) = v5 - x x+3 (d) q(x) =7247x+12 9-x', x