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Math 1513 - College Algebra Exam #5 - 2016.11.16 Solutions 1. Let f (x) = 2x − 1 . x+3 (a) State the domain and range of f (x). The domain will be all numbers except where the denominator is zero, thus (−∞, −3) ∪ (−3, ∞). The range is all real numbers except y = 2 – (−∞, 2) ∪ (2∞). (b) Graph the function f (x). (c) Algebraically compute f −1 (x). We replace x with y and y with x and then solve for y: x= 2y − 1 → x(y + 3) = 2y − 1 y+3 → xy + 3x = 2y − 1 → 1 + 3x = 2y − xy → 1 + 3x = (2 − x)y 1 + 3x →y= 2−x (d) Graph the function f −1 (x). 1 2 (e) Verify that (f ◦ f −1 )(x) = x. (f ◦ f −1 )(x) = f = 1 + 3x 2−x 2 1+3x 2−x − 1 1+3x 2−x +3 2(1 + 3x) − 1(2 − x) (1 + 3x) + 3(2 − x) 2 + 6x − 2 + x = 1 + 3x + 6 − 3x 7x = 7 =x = 2. Sketch a graph of the function g(x) = 3−x+1 − 2. We start with y = 3x , then shift to the right one unit: y = 3x−1 . Next, we reflect about the y−axis to get y = 3−(x−1) . Finally, we shift the graph two units down to get g(x). The dashed blue curve below is the original graph y = 3x , and the solid black line is our function g(x). 3. Determine the domain of h(x) = log42 −x2 (x − 1)(x + 1)3 . We need to find all the values of the argument p(x) = −x2 (x − 1)(x + 1)3 such that p(x) > 0. The real line gets broken into subintervals (−∞, −1) ∪ (−1, 0) ∪ (0, 1) ∪ (1, ∞) and we test each interval. The positive intervals are (−1, 0) and (0, 1). Thus the domain of h(x) is (−1, 0) ∪ (0, 1). 4. Graph the function r(x) = log2 (2x − 1) + 1. First we rewrite the function to get a better idea: y = log2 (2x − 1) + 1 1 +1 = log2 2 x − 2 1 = log2 (2) + log2 x − +1 2 1 + 1 + log2 (2) = log2 x − 2 3 So now we recognize that this is the graph of log2 (x) shifted to the right 12 units and then up 1 + log2 (2) units. The dashed blue curve below is the original graph y = log2 (x), and the solid black line is our function r(x). 5. Use properties of logarithms to write the following expression as sums or differences of simple logarithmic terms. s ! 2 4 3 (x + 1) (y − 2) (z + 1) log3 (a + 5)7 (b + 1)2 (c − 4)4 We use the power rule first: s log3 3 (x + 1)2 (y − 2)4 (z + 1) (a + 5)7 (b + 1)2 (c − 4)4 ! 1 = log3 3 (x + 1)2 (y − 2)4 (z + 1) (a + 5)7 (b + 1)2 (c − 4)4 Next we use the product and quotient rules: 1 (x + 1)2 (y − 2)4 (z + 1) 1 = log3 ((x + 1)2 ) + log3 ((y − 2)4 ) + log3 (z + 1) log3 7 2 4 3 (a + 5) (b + 1) (c − 4) 3 − log3 ((a + 5)7 ) − log3 ((b + 1)2 ) − log3 ((c − 4)4 ) Lastly, we use the power rule once more: s ! 2 4 1 3 (x + 1) (y − 2) (z + 1) = [2 log3 (x + 1) + 4 log3 (y − 2) + log3 (z + 1) log3 7 2 4 (a + 5) (b + 1) (c − 4) 3 −7 log3 (a + 5) − 2 log3 (b + 1) − 4 log3 (c − 4)] 6. Solve for x in the following equation: x+2 1 = 2 · 53x−1 3 We first take log of both sides (pick your favourite base): x+2 ! 1 log = log 2 · 53x−1 3 Simplifying using the product and power rules gives 1 = log(2) + (3x − 1) log(5) (x + 2) log 3 Next we multiply everything out: 1 1 x + 2 log = log(2) + 3 log(5)x − log(5) log 3 3 Everything with x’s in it gets moved to the left, and the rest to the right. 1 1 log x − 3 log(5)x = log(2) − log(5) − 2 log 3 3 4 Now we factor x out of the left: 1 1 − 3 log(5) x = log(2) − log(5) − 2 log log 3 3 And lastly, we divide by the expression multiplied by x on the left to finally solve for x. log(2) − log(5) − 2 log 31 x= log 31 − 3 log(5) 7. Solve for x in the following equation: log(1 − x) + log(x) = log(x − 4) This equation can have no solution, as the domains of each individual logarithmic function have an empty intersection. If we do try, we get the following: log((1 − x)x) = log(x − 4) Using 10 as a base for each side gives (1 − x)x = x − 4 Solving for x gives x2 − 4 = 0 or x = ±2. Neither of those are valid solutions. 8. Solve for x in the following equation: log3 (x − 2) = 1 − log3 (2x + 1) We move the terms with x in them to the right log3 (x − 2) + log3 (2x + 1) = 1 and then use the product property of logarithms log3 ((x − 2)(2x + 1)) = 1 Using 3 as a base for each side (x − 2)(2x + 1) = 3 Multiplying the left side out gives 2x2 − 3x − 2 = 3 Setting equal to zero: 2x2 − 3x − 5 = 0 Factoring gives The solutions are x = −1 and x = 5 2. (x + 1)(2x − 5) = 0. Only the x = 25 solution is valid, as x = −1 is not in the domain of log3 (2x + 1).