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Material Type: Exam; Professor: Moorhouse; Class: Calculus I; Subject: Mathematics; University: Colgate University; Term: Fall 2008;
Typology: Exams
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Math 111 – Calculus 1 Practice Problems for Exam 3 Fall 2008
(a) local maxima: f(-1)=27 and f(3)=27 minima: f(1)=- (b) absolute maximum: 27 minimum: -
(a) limx→ 0 sinx^ x 3 −x
lim x→ 0
sin x − x x^3
L =′H lim x→ 0
cos x − 1 3 x^2
L′H = lim x→ 0
sin x 6 x
L =′H lim x→ 0
cos x 6 =
(b) lim t→ 0 (sec x − tan x) = 1
(c) lim x→∞ x^2 e−x
lim x→∞
x^2 ex
L′H = lim x→∞
2 x ex
L′H = lim x→∞
ex = 0
(d) lim x→ 0 x^2 ln x^2
lim x→ 0
ln x^2 x−^2
L =′H lim x→ 0
1 x^2 2 x − 2 x−^3
= lim x→ 0
x
x^3 − 2
(e) lim t→∞ t − ln t
Note that t − ln t = ln et−ln^ t^ = ln e
t t and by L’Hospital’s Rule
lim t→∞
et t
= lim t→∞ et
so
lim t→∞ t − ln t = lim t→∞ ln
et t = lim t→∞ ln et = lim t→∞
t = ∞
(f) lim x→ 0
2 x^ − 1 x
lim x→ 0
2 x^ − 1 x
L = lim′H x→ 0
2 x^ ln 2 1
= ln 2
(g) lim x→ 0
arcsin x x
lim x→ 0
arcsin x x
L′H = lim x→ 0
√^1 1 −x^2 1 = 1
(h) lim x→ 1
(ln x)^2 x
(a) Find the x and y intercepts of f.
x−intercept: none
y−intercept: y = 2
(b) Find the vertical and horizontal asymptotes of f. Vertical asymptote: none
Horizontal asymptote: y = 0 this asymptote is only in the negative direction. As x → ∞, f (x) → ∞.
(c) Find the intervals on which f is increasing or decreasing.
Decreasing: (0, 4) Increasing: (−∞, 0), (4, ∞) (d) Find all the points at which f has a local maximum or a local minimum. Local maximum at (0, 2)
Local minimum at (4, 10 e^4 )
(e) Determine the intervals on which f is concave up or concave down, and find the inflection points. Concave down: (−∞, 1 −
Concave up: (1 −
Inflection points occur at x = 1 −
5 and x = 1 +
(f) Sketch the graph of f. Be sure to label all the x and y values of the intercepts, local maxima, local minima, inflection points, and asymptotes.
We first notice that f (1) = 0 so f has a root at x = 1. Now, suppose for the sake of contradiction that there were two roots, a and b, then f (a) = f (b) = 0. Since f is differentiable on all of R, by Rolle’s Theorem there must exist some c between a and b such that f ′(c) = 0. But notice that f ′(x) = − 1 − 3 x^2 ≤ −1 so that f ′(x) cannot equal zero for any x. Thus, we have a contradiction and our assumption must be false. I.e. there cannot be two roots.
(b) Let f (x) be a function that is differentiable for all x. Suppose that f (0) = −3, and f ′(x) ≤ 5 for all values of x. How large can f (2) possibly be? (You must use the Mean Value justify your answer, even if you can do it in your head.)
Since f is differentiable, by the Mean Value Theorem there must exist some c between 0 and 2 where f ′(c) = f^ (2) 2 −−f 0 (0)= f^ (2)+3 2. But we know that f ′(c) ≤ 5 so we have f (2) + 3 2
Solving this inequality for f (2) we find that f (2) ≤ 7.
See the answers in the back of your text.
(a) For what values of x is g(x) increasing? For what values of x is g(x) decreasing? (b) For what values of x (if any) does g(x) have a vertical tangent or a vertical cusp? (c) For what values of x is g(x) concave up? For what values of x is g(x) concave down? (d) Sketch a graph of f assuming that f (0) = 0.
If we let b be the base and h the height of the rectangle, then using similar triangles we find that h =
3(2 − b/2) Substituting this into the area formula, we get
A = bh =
3(2b − b^2 /2).
Note that this function is only defined for 0 ≤ b ≤ 2
Differentiating and solving gives b = 2 and we can see that this must be a maxi- mum since at b = 0 and at b = 2
3 give A = 0.
(b) A farmer wants to build a rectangular pen and then split it into 3 pens of equal size. If the farmer has 800 feet of fencing, what is the maximum possible total area that can be enclosed? What are the dimensions of each pen in that case?