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QUIZ 4 – detailed solutions
1) What is the y‐intercept of 5x ‐ 24 = 10 . This equation represents a line. Think y = mx + b (slope ‐ intercept form). Solve the equation for y as follows 5x ‐ 2y = 10 ‐ 2y = ‐5x + 10 Divide by ‐2 y = 5/2 x ‐ 5 Now, the constant term represents the y‐intercept. So our answer is ‐5. 2) What is the slope of 5x ‐ 2y = 10. This equation represents a line. Think y = mx + b (slope ‐ intercept form). Solve the equation for y as follows 5x ‐ 2y = 10 ‐ 2y = ‐5x + 10 Divide by ‐2 y = 5/2 x ‐ 5 Now, the coefficient of the x term represents the slope. So our answer is 5/2. 3) Evaluate lim_{x rightarrow infty} x 3 as x gets large in a positive direction x 3 (x cubed) does also. So the answer here is positive infinity. 4) Evaluate lim_{x rightarrow ‐ infty} x 3 . As x becomes a very large negative number, x 3 (x cubed) will become a very large negative number since 3 negatives multiplied together produce a negative. 5) Evaluate lim_{x rightarrow infty} 3 x . As x becomes very large, 3 to a very large number will be a very large number. So the answer here is infinity. 6) Evaluate lim_{x rightarrow ‐ infty} 3 x . As x becomes a very large negative number we must remember to evaluate an expression with a negative exponent. It is best to move the term to the denominator where the exponent is then positive. That is, evaluate lim_{x rightarrow infty} 1 / 3 x . Now, 3 x will become very large as x moves towards infinity. So 1 divided by this very large number will be a very small number. So, lim_{x rightarrow ‐ infty} 3 x is 0. 7) If f(x) = 3x ‐ 2 and g(x) = 1/x+2, what is f(g(x))? This problem requires knowledge of what is commonly referred to as the "composite function". It is noteworthy that this is often notated f O g(x) or f(g(x)). All that is requires is x is replaced by g(x) in the f(x) function. So, f(g(x)) = 3( 1/x+2 ) ‐ 2 = 3 / x+2 ‐ 2 Next, we create a common denominator of x + 2 by multiplying the constant 2 in both its numerator and denominator (1) by x + 2. Doing so, = 3 / x+2 ‐ 2(x+2) / x+2 Creating the single fraction and being cautious distributing the ‐2 factor we get = (3‐2x‐4) /( x+2) and finally combining f(g(x)) = (‐2x‐1) /( x+2) 8) If f(x) = 3x ‐ 2 and g(x) = 1 / x+2, what is g(f(x))? This problem requires knowledge of what is commonly refered to as the "composite function". It is noteworthy that this is often notated g O f(x) or g(f(x)). All that is requries is x is replaced by f(x) in the g(x) function. So, g(f(x)) = 1 / 3x‐2+2 = 1 / 3x 9) Find an equation for the line that passes through the points (0,1) and (2,‐1). Since two points determine a line, there will be one unique solution to this problem. To write the equation of a line we employ the slope‐intercept formula. (y = mx + b). The y‐intercept (b) is actually given to us. By requiring the line pass through the point (0,1) we know the y‐intercept (b) must be 1, since (0,1) lies on the y‐axis. Next, to find the slope (m) of the line we remember that the slope of a line is the ratio (fraction) between the change in y‐values divided by the change in x‐values. This is described by the formula m = y 2 – y 1 / x 2 – x 1 Plugging the x and y values into their respective positions in the above formula we get m = 1 – ‐1 / 0 – 2 simplifying m = 2 / ‐2 = ‐1 Now, since m = ‐1 and b = 1, the equation of the line must be y = ‐1x + 1 or y = ‐x + 1 10) Describe in words, three key features of the graph of the function f(x) = 2 x . We begin by identifying the y‐intercept of the equation. The y‐intercept is found by plugging in an x‐value of 0. By doing so f(x) = 2 0 = 1 So we know the y‐intercept is 1. Next, we evaluate the function as x goes to infinity. lim_{x rightarrow + infty} 2 x . Clearly, as x gets infinity large, 2 x gets infinity large. Next, we evaluate the function as x goes to negative infinity. lim_{x rightarrow + infty} 2 x . As x get large negatively we consider 1 / 2 x as x gets large, but since x has been moved to the denominator it is now a positive number. Consequently we are looking at a fraction whose numerator is a constant 1 and who’s denominator is growing increasingly larger as a result, the whole fraction is getting smaller and smaller or in short approaching 0. It should be noted that f(x) will never be negative. That is, f(x) will always lie above the x‐axis. B. Smith CUMATH Clarkson University Potsdam NY 13699