Stitz-Zeager_College_Algebra_e-book

# 5 and that the half life10 of iodine 131 is

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: axis. 362 Exponential and Logarithmic Functions (+) 0 (−) 0 (+) −1 4 y = r(x) = 2x 2 −3x − 16 x e 2. The ﬁrst step we need to take to solve ex −4 ≤ 3 is to get 0 on one side of the inequality. To that end, we subtract 3 from both sides and get a common denominator ex ex − 4 ≤3 ex −3 ≤ 0 −4 3 (ex − 4) ex − ≤ 0 Common denomintors. ex − 4 ex − 4 12 − 2ex ≤0 ex − 4 ex x −2 We set r(x) = 12x −e and we note that r is undeﬁned when its denominator ex − 4 = 0, or e4 when ex = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4), ∞). To ﬁnd the zeros of r, we solve r(x) = 0 and obtain 12 − 2ex = 0. Solving for ex , we ﬁnd ex = 6, or x = ln(6). When we build our sign diagram, ﬁnding test values may be a little tricky since we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4 which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the sign of r (ln(3)) may be very unsettling, remember that eln(3) = 3, so r (ln(3)) = 12 − 2eln(3) 12 − 2(3) = −6 = ln(3...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern