Stitz-Zeager_College_Algebra_e-book

12 13 at which point it would be more toast than

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Unformatted text preview: 2ex (n) 70 + 90e−0.1t = 75 (i) 73+7x = 34−2x 1 (x+5) 2 3(x−1) 1 2x 9 28e−6x (g) 9 · 37x = =3 (o) (p) 150 = 75 1+29e−0.8t x 4 25 5 = 10 2. Solve the following inequalities analytically. (a) ex > 53 (b) 1000 1 + (c) 3 2(x −x) 4x ≥ 10 5 150 ≤ 130 1+29e−0.8t −0.1t ≤ 75 70 + 90e (d) 25 0.06 12t 12 (e) ≥ 3000 <1 (f) 3. Use your calculator to help you solve the following equations and inequalities. (a) ex < x3 − x (b) 2x = x2 √ (c) e x = x + 1 (d) e−x − xe−x ≥ 0 (e) 3(x−1) < 2x (f) ex = ln(x) + 5 4. Since f (x) = ln(x) is a strictly increasing function, if 0 < a < b then ln(a) < ln(b). Use this fact to solve the inequality e(3x−1) > 6 without a sign diagram. 5. Compute the inverse of f (x) = ex − e−x . State the domain and range of both f and f −1 . 2 6. In Example 6.3.4, we found that the inverse of f (x) = 5ex was f −1 (x) = ln ex + 1 x 5−x but we left a few loose ends for you to tie up. (a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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