This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Spring 2011 Math 151 Exam III Version A Solutions 1. D Since the numerator and denomina- tor both approach 0, use L’Hospital’s Rule: lim x → e x- cos x- 2 x x 2- 2 x = lim x → e x + sin x- 2 2 x- 2 = 1 2 . 2. E Use properties of logarithms: ln( x 2 + x ) = ln( x +4) , so x 2 + x = x +4 , x 2- 4 = 0 , which yields x = 2 or x =- 2 . Since the domain of the left-hand side of the original equation is x > , the only solution is x = 2 . 3. A f ′ positive means f is increasing, and f ′ decreasing means f ′′ is negative, so f is con- cave down. The only graph increasing and concave down is graph A . 4. B The original function f is decreasing when f ′ is negative, which occurs when x ∈ ( a , c ) ∪ ( e , ∞ ) . 5. B f has a critical value when f ′ = 0 , namely when x = a, c, e . Using the signs of the derivative f ′ , we find that f is increasing for x ∈ (-∞ , a ) ∪ ( c, e ) and decreasing for x ∈ ( a, c ) ∪ ( e, ∞ ) . Therefore, f has a local minimum only when x = c ....
View Full Document
This note was uploaded on 02/13/2012 for the course MATH 151 taught by Professor Artbelmonte during the Spring '06 term at Texas A&M.
- Spring '06