Section 2.3 Properties of Functions

22 Section 2.3 Properties of Functions In this section, we will explore different properties of functions that will allow us to obtain the graph of the function more quickly. Objective #1 Determining Even and Odd Functions from a Graph. In mathematics, we use the words even and odd to describe the symmetry of the graph of a function. Let's begin with some definitions. Even and Odd Functions and Symmetry 1) A function is even if, for every ( x, y) (x, y) number x in the domain, the number x is also in the domain and f( x) = f(x) = y. This means that the function is symmetric with respect to the y-axis if and only if it is even. 2) A function is odd if, for every (x, y) number x in the domain, the number x is also in the domain and f( x) = f(x) = y. This means that the function is symmetric with respect to the origin if and only if it is odd. ( x, y) Determine if the following functions are even, odd, or neither: Ex. 1a Ex. 1b Ex. 1c 4 3 2 1 - -4-3 -2-1 1 2 3 4-2 -3-4 - 6 4 3 2 1-4 -3-2 -1-1 1 2 3 4-2

a) Since the graph is not symmetric with respect to the y-axis and not symmetric with respect to the origin, the function is neither even nor odd. b) Since the graph is symmetric with respect to the y-axis, the function is even. c) Since the graph is symmetric with respect to the origin, the function is odd. 23 Objective #2: Determining if a Function is Even, Odd, or Neither Algebraically. To determine if a function is even or odd algebraically, we replace x by x and simplify to see if we get f( x) = f(x) (even; original function) or f( x) = f(x) (odd; opposite of the original function). Determine if the following functions are even, odd, or neither: Ex. 2a f(x) = 4x 2 1 Ex.2b g(x) = x x 2 + 1 3 Ex. 2c h(x) = x + x We will begin by plugging x in for x and then try to relate it back to the original function: a) f( x) = 4( x) 2 1 = ( x) 2 4x2 1 = f(x). x 2 Since f( x) = f(x), then f is an even function. b) g( x) = x + 1 = x + 1 = ( x 1) g(x) or g(x). Since g( x) g(x) and g( x) g(x), then g in neither an even or odd function. 3 c) h( x) = x + ( x) 3 = x x 3 = ( x + x ) = h(x) Since h( x) = h(x), then h is an odd function. If we were to use a graphing utility, we can see that the graph of f is symmetric with respect to the y-axis and the graph of h is symmetric with respect to the origin. The graph of g is neither symmetric to the origin nor to the y-axis.

24 f is symmetric with g is neither symmetric h is symmetric respect to the y-axis. with respect to the y-axis with respect to nor to the origin. the origin Objective #3 Determining when a Function is Increasing, Decreasing, or Neither. If we were to plot the miles per gallon a car was achieving versus its speed, we would find that the faster the car travels, the better gas mileage it would get until the car s speed was near mph. Once the car s speed exceeded that point, the gas mileage would start to decrease as the car started going faster. The optimal speed range would be between and 6 mph. That is where the car is most fuel-efficient. Let s take a closer look at the graph: 3 2 Fuel Efficiency vs. Speed Gas Mileage (MPG) 2 1 1 1 2 3 4 6 7 8 9 Speed of the Car (MPH)

2 Notice that as the speed increases from mph to about 2 mph, the fuel efficiency increases. Thus, the function is increasing on the interval (, 2). When the speed is between 2 mph and about 7 mph, the fuel efficiency stays constant. Thus, the function is constant on the interval (2, 7). Finally, when the speed exceeds 7 mph, the fuel efficiency decreases. Hence, the function is decreasing on the interval (7, 8). Intervals over which a function is Increasing, Decreasing, or Constant Let I be an open interval in the domain of the function f. Then, 1) f is increasing on I if f(a) < f(b) for all a < b in I. (a function is increasing on an interval if it rises or goes uphill as you move from left to right. 2) f is decreasing on I if f(a) > f(b) for all a < b in I. (a function is decreasing on an interval if it falls or goes downhill as you move from left to right. 3) f is constant on I if f(a) = f(b) for all a and b in I. a function is constant on an interval if it stays level or flat as you move from left to right. Determine where the following function is a) increasing, b) decreasing, and c) constant: Ex. 3 Cost in Dollars 1 9 8 7 6 4 3 2 1 Cost to Drive to Houston vs. Speed 2 4 6 8 1 Speed of the Car (MPH)

a) The function rises as we move from about 8 mph to 8 mph, thus the function is increasing on (8, 8). b) The function falls as we move from mph to about 8 mph, thus the function is decreasing on (, 8). c) The function is constant nowhere. Ex. 4 Ex. 26 4 3 2 1-4 -3-2 -1-1 1 2 3 4-2 -3-4 4 3 2 1 - -4-3 -2-1 -1 1 2 3 4-2 -3-4 - 4a) The function rises as we move from -1 to and 2 to, thus the function is increasing on ( 1, ) U (2, ). 4b) The function falls as we move from to 1 and to 2, thus the function is decreasing on (, 1) U (, 2). 4c) The function is constant nowhere. a) The function rises as we move from - to, thus the function is increasing on (, ). b) The function falls as we move from 3 to, thus the function is decreasing on (3, ). c) The function stays the same as we move from to 3, thus the function is constant on (, 3).

27 Objective #4: Finding Local Maximum & Minimum Values of a Function. If we look at the graph from example #4, we can see that the graph has two "valleys". The lowest points in these "valleys" occur at x = 1 and at x = 2. We can say that f( 1) and f(2) are local minimum values (relative minimum). Similarly, the graph has one "top of a hill" point. The "top of the hill" point occurs at x =. We can then say that f() is a local maximum value (relative maximum). 4 3 2 1-4 -3-2 -1 1 2 3 4-1 "valleys" Definition of local extrema A function f has a local (relative) maximum at x = a if f(a) f(x) for all x in an interval c < x < d containing a. A function f has a local (relative) minimum at x = b if f(b) f(x) for all x in an interval c < x < d containing b. In other words, a function has a local maximum at x = a if y = f(a) is the "highest" y-value in a small neighborhood containing a and a function has a local minimum at x = b if y = f(b) is the "lowest" y-value in a small neighborhood containing b. Notice that with our definition, the local maxima (plural of maximum) and minima (plural of minimum) occur when the function is changing from increasing to decreasing or vice-versa. Find the local maxima and minima values of the following functions: Ex. 6a Ex. 6b -2-3 -4 "top of a hill"

28 a) There are two peaks at x = 4 and x 4. and one valley at x = 1. Thus, the function has two local maxima, one of f( 4) = 6 at x = 4 and another of f(4.) = 3 at x = 4.. The function has a local minimum of f(1) = 1 at x = 1. b) There is a peak at x 3. and one valley at x 3.. Thus, the function has a local maximum of f(3.) = 2 at x = 3. and a local minimum of f( 3.) = 2 at x = 3.. Objective # Finding Absolute Extrema of a Function. In example #4, the function had to local minima; one of 4 at x = 1 and the other of 32 at x = 2. The second local minimum of 32 at x = 2 was the lowest point of the entire graph. We call such a point the absolute minimum of the function. Since the function in example #4 increasing without bound as x get large both positively and negative, the graph has no absolute maximum. In example #6a, the function had to local maxima; one of 6 at x = 4 and the other of 3 at x = 4.. The first local maximum of 6 at x = 4 was the highest point of the entire graph. We call such a point the absolute maximum of the function. Since the function in example #6a decreasing without bound as x get large both positively and negative, the graph has no absolute minimum. Definition of absolute extrema A function f has an absolute maximum at x = a if f(a) f(x) for all x in the domain of f. A function f has an absolute minimum at x = b if f(b) f(x) for all x in domain of f. Find the absolute extrema of the following functions: Ex. 7a Ex. 7b 6 4 2-4 - 3-2 - 1-2 1 2 3 4-4 - 6-8 - 1-12 - 14

Ex. 8a a) The lowest point on the graph is ( 6, 2) so the function has an absolute minimum of 2 at x = 6. The highest point on the graph is ( 2, 8), so the function has an absolute maximum of 8 at x = 2. b) The lowest point on the graph is ( 3, 14) so the function has an absolute minimum of 14 at x = 3. Since the function increases without bound for large positive and negative x, there is no absolute maximum. Ex. 8b 29 a) The lowest point on the graph seams to be near (3, 1), but x = 3 is excluded from the domain since there is a hole at (3, 1). Hence there is no absolute minimum. Since the function increases without bound for large positive and negative x, there is no absolute maximum either. b) The lowest point on the graph is every point between ( 1, 1) and (2, 1) inclusively, so all of those points are absolute maxima. The highest point on the graph is ( 3, 9), so the function has an absolute maximum of 9 at x = 3. In Calculus, you will study an important theorem called the Extreme Value Theorem which applies to functions that are "continuous" (i.e., a function is continuous at a point c if you can draw the graph through c without lifting a pencil). The theorem states that if a function is "continuous" on a closed interval [a, b], then f has an absolute maximum and minimum on [a, b].

3 Objective #6: Finding the Absolute Extrema in an Application: Ex. 9 An elementary school plans to build a rectangular playground that is 4,9 square meters in area. The playground is to be surrounded by a fence. Express the length of the fencing as a function of the length of one of the sides of the playground, draw the graph, and find the dimensions of the playground requiring the least amount of fencing. Let L = the length of the playground & w = the width of the playground The perimeter of the playground is: P = 2L + 2w and the area is A = Lw = 49. Solving the area equation for L yields: L = 49. Substituting this answer into the perimeter equation yields: w P = 2L + 2w = 2( 49 w Thus, our function is P(w) = 98 w ) + 2w = 98 w + 2w. + 2w. The domain is w > As an aid to graphing the function, let s generate a table of values, plot the points, and draw a smooth curve through the points: w 1 2 3 4 6 7 P(w) 1 3 387 32 296 283 28 w 8 9 1 11 12 13 14 P(w) 282. 289 298 39 322 33 3

From the graph, it appears that the dimensions should be 7 feet by 7 feet in order to use the minimum amount of fencing. 31 Objective #7: Average Rate of Change. In order to calculate the slope of a line, we took the vertical change (y 2 y 1 ) divided by the horizontal change (x 2 x 1 ). Thus, the slope gives us the average rate of change in y-units compared to a unit of x. To illustrate how this works, consider the following example: Calculate the slope: Ex. 1 Leroy started his driving trip 8 miles north of Austin and headed north at 6 mph. After every hour, Leroy plotted the distance he was from Austin versus the time had been on the road. The graph he created is to the right. Calculate the slope and interpret what it represents. 4 36 32 28 24 2 16 12 8 4 Distance from Austin (miles) (3, 27) (2, 21) (1, 14) Hours Driving 1 2 3 4 6 7 8 9 1 We first pick two distinct points on the graph: (2, 21) & (3, 27). Calculating the slope, we get: m = y 2 y 1 x 2 x 1 = 27 21 3 2 = 6 1 = 6. Thus, the slope is the average rate of change of the distance travelled compared to one hour of time or more simply, the average speed. For any two distinct points on a curve, we can find the average rate of change from the first point to the second point by drawing a straight line (secant line) through those points and calculating the slope of that line. This will give us the average rate of change between those two points. Average Rate of Change The average rate of change in the function y = f(x) from x = a to x = b is average rate of change = change in y change in x f(b) f(a) = b a

Average rate of change is the slope of the secant line between the points (a, f(a)) and (b, f(b)) 32 f(x) f(b) Secant line (b, f(b)) f(a) (a, f(a)) a Determine the average rate of change of the function between the given values of the variable. Then find the equation of the secant line: Ex. 11a f(x) = 4x 2 x 3 ; x = 2, x = 4 Ex. 11b g(x) = x 2 ; x = 3, x = a) f(4) = 4(4) 2 (4) 3 = and f( 2) = 4( 2) 2 ( 2) 3 = 24 Average rate of change = b f(b) f(a) b a = f(4) f( 2) 4 ( 2) = 24 4+2 = 24 6 = 4. This also is the slope of the secant line. Using m = 4 and (x 1,y 1 ) = (4, ), we can find the equation of the secant line: y y 1 = m(x x 1 ) y = 4(x 4) y = 4x + 16 b) g() = () 2 = 2 and g(3) = (3) 2 = 4 Average rate of change = g(b) g(a) b a = g() g(3) (3) = 2 4 2 = 16 2 = 8. This also is the slope of the secant line. Using m = 8 and (x 1,y 1 ) = (3, 4), we can find the equation of the secant line: y y 1 = m(x x 1 ) y 4 = 8(x 3) y 4 = 8x 24 y = 8x 2