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MATH: PRE-CALCULUS Section IV DePaul Math Placement Test

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[object Object],Evaluating a function simply means finding f(x) at some specific value x. Here you will have to evaluate a function at some particular constant. Take a look at the following example: If f(x) = x 2 – 3 , what is f(5)? Evaluating a function at a constant involves nothing more than substituting the constant into the definition of the function. In this case, substitute 5 for x: f(5) = 5 2 -3 = 22 In the Placement test there might be questions based on how to evaluate a function at a variable rather than a constant. Q. If f(x) = 3x / 4–x , what is f(x + 1)? A. To solve problems of this sort, follow the same method you used for evaluating a function at a constant: substitute the variable into the equation. To solve the sample question, substitute x + 1) for x in the definition of the function: f(x + 1)= f(x+1) = f(x+1) =

[object Object],Functions can be added, subtracted, multiplied, and divided like any other quantity. There are a few rules that make these operations easier. For any two functions f(x) and g(x): Rule Example Addition (f+g)(x)=f(x) +g(x) If f(x) = sin x, and g(x) = cos x: (f + g)(x) = sin x + cos x Subtraction (f-g)(x)=f(x) -g(x) If f(x) = x 2 + 5, and g(x) = x 2 + 2x + 1: (f – g)(x) = x 2 + 5 – x 2 – 2x – 1 = –2x + 4 Multiplication (f X g)(x)=f(x) X g(x) If f(x) = x, and g(x) = x 3 + 8: (f+g)(x)=x(x 3 +8)= x 4 +8x Division (f ÷ g)(x)=f(x) If g(x) 0 If f(x) = 2 x and g(x) = 2 x 2 (f g)(x)=f(x) g(x) = 2x/x 2 = 1/x. Here you have to be aware of possible situations in which you inadvertently divide by zero. Since division by zero is not allowed , you should remember that any time you are dividing functions, like f (x) ⁄ g(x) , the result of the function is undefined.

[object Object],Recognizing that these two situations cause the function to be undefined is the key to finding any restriction on the function’s domain . Once you’ve discovered where the likely problem spots are, you can usually find the values to be eliminated from the domain easily. Q. What is the domain of f(x)= ? A. In this question , f(x) has variables in its denominator, which should be a red flag that alerts you to the possibility of division by zero. We may need to restrict the function’s domain to ensure that division by zero does not occur. To find the values of x that cause the denominator to equal zero, set up an equation and factor the quadratic: x 2 + 5x + 6 = (x + 2)(x + 3) = 0. For x = {–2, –3}, the denominator is zero and f(x) is undefined. Since it is defined for all other real numbers, the domain of f(x) is the set of all real numbers x such that x ≠ –2, –3 . This can also be written as {x: x ≠ –2, –3}.

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[object Object],What is the inverse of the function f(x) = ? First, replace f(x) with y. Then switch the places of x and y, and solve for y. x = x 2 = 5x 2 = 2y 2 – 3 5x 2 + 3= 2y 2 (5/2)x 2 + (3/2) =y 2 y= f –1 (x)=

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[object Object],Here’s another question: Q .What is the inverse of f(x) = 2|x – 1|, and is it a function? A. Replace x with y and solve for y: x = 2|y – 1| 1 x =|y-1| 2 Now, since you’re dealing with an absolute value, split the equations : y-1 =1/2 x or 1-y=1/2x Therefore, y=1/2x +1 or y = -1/2x + 1 The inverse of f(x) is this set of two equations. As you can see, for every value of x except 0, the inverse of the function assigns two values of y. Consequently, f –1 (x) is not a function.

[object Object],A compound function is a function that operates on another function . A compound function is written as nested functions, in the form f(g(x)). To evaluate a compound function, first evaluate the internal function , g(x). Next, evaluate the outer function at the result of g(x). Work with the inner parentheses first and then the outer ones, just as in any other algebraic expression. Here’s an example: Suppose h(x) = x 2 + 2x and j(x) = | (x /4) + 2|. What is j(h(4))? To evaluate this compound function, first evaluate h(4): h(4) =4 2 +2(4) =24 Now plug 24 into the definition of j: j(24)=| 24/4+2| j(24)=8 It is important that you pay attention to the order in which you evaluate the compound function. Always evaluate the inner function first . For example, if we had evaluated j(x) b efore h(x) in the above question, you would get a completely different answer: h(j(4))=h(|4/4 +2|) =h(|1+2|) =h(3) =15

[object Object],Here Suppose f(x) = 3x + 1 and g(x) = . What is g(f(x))? When you are not given a constant at which to evaluate a compound function, you should simply substitute the definition of f ( x ) as the input to g ( x ). This situation is exactly the same as a regular equation being evaluated at a variable rather than a constant. g(f(x)) =g(3x+1) = =

[object Object],The range and domain of a function are easy enough to see in their graphs. The domain is the set of all x -values for which the function is defined. The range is the set of all y -values for which the function is defined. To find the domain and range of a graph, just look at which x - and y -values the graph includes. Certain kinds of graphs have specific ranges and domains that are visible in their graphs. A line whose slope is not 0 (a horizontal line) or undefined (a vertical line) has the set of real numbers as its domain and range. Since a line, by definition, extends infinitely in both directions, it passes through all possible values of x and y :

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[object Object], The vertical line test makes sense because the definition of a function requires that any x -value have only one y -value. A vertical line has the same x -value along the entire line; if it intersects the graph more than once, then the graph has more than one y -value associated with that x -value. Using the vertical line test, check to see that the three graphs below are functions. The next three graphs are not functions. In each graph, a strategically placed vertical line (depicted by the dashed line) will intersect the graph more than once.

[object Object], While most of the function questions on the Math placement test involves analysis and manipulation of the functions themselves, you will sometimes be asked a question about the graph of a function. A common question of this type asks you to match a function’s graph to its definition. Understanding the next few topics will help prepare you for questions relating to functions and their graphs. Identifying Whether a Graph is a Function For Placement test, it is important to be able to determine if a given graph is indeed a function. A foolproof way to do this is to use the vertical line test: if a vertical line intersects a graph more than once, then the graph is not a function. The vertical line test makes sense because the definition of a function requires that any x - value have only one y - value. A vertical line has the same x -value along the entire line; if it intersects the graph more than once, then the graph has more than one y - value associated with that x - value. Using the vertical line test, check to see that the three graphs below are functions. Continued next page.

[object Object],Trigonometric functions have various domains and ranges depending on the function. Sine, for example, has the real numbers for its domain and {–1, 1} for its range. A more detailed breakdown of the domains and ranges for the various trigonometric functions can be found in the Trigonometry chapter. Some functions have limited domains and ranges that cannot be categorized simply, but are still obvious to see. By looking at the graph, you can see that the function below has domain {3, ∞} and range {–∞, –1}. A Trigonometric function graph A limited domain and range graph

[object Object],Logarithms are closely related to exponents and roots. A logarithm is the power to which you must raise a given number, called the base, to equal another number. For example, log 2 8 = 3 because 2 3 = 8 . In this case , 2 is the base and 3 is the exponent. Having defined logarithms in a sentence, let’s show one symbolically. The next two equations are equivalent: log a x = b as a b = x log 10 1000 = 3 because 10 3 =1000 log 4 ¼ = 2 because (1/2) 2 = ¼ For example, log 4 16 = 2 because 4 2 = 16 and = 4. You should now be able to see why the three topics of exponents, roots, and logarithms are often linked together. Each method provides a way to isolate one of the three variables in these types of equations. In the example above, a is the base, b is the exponent, and x is the product. Finding the root, logarithm, and exponent isolates these values, respectively.

[object Object],Exponential Functions - The exponential function is a function in mathematics. The application of this function to a value x is written as exp(x) . Equivalently, this can be written in the form e x , where e is a mathematical constant, the base of the natural logarithm, which equals approximately 2.718281828. Natural Logarithms- The natural logarithm , the logarithm to the base e, where e is an irrational constant approximately equal to 2.718281828. In simple terms, the natural logarithm ( ln) of a number x is the power to which e would have to be raised to equal x — for example the natural log of e itself is 1 because e 1 = e, while the natural logarithm of 1 would be 0 , since e 0 = 1 NOTE- All the laws of the of logarithms with base 10 can be applied to all natural logarithm The relationship between exponential function a x (called the exponential function with base a ) is defined using the natural logarithms as follows: e ln(x) = x if x>0 ln(e x ) = x. And, a x =( e lna ) x = e x ln a Now, test your knowledge of the topics discussed by clicking on the sample test links given below.

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