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Math 111 โ Calculus 1 Solutions to Practice Problems for Exam 1 Fall 2008
- Find the inverse of the function f (x) = 1+ 1 โxx. Answer: f โ^1 (x) = x1+โ^1 x
- Find the exponential function whose graph passes through the points (1, 2) and (โ 1 , 92 ). Answer: y = 3(^23 )x
- Given
f (x) = ln (x โ 2) + 3.
(a) Sketch a graph of f. The sketch does not have to be perfect, but it should have the correct shape, and the line x = 2 should be clearly indicated.
Answer: Your function should look like ln x shifted to the right 2 units and up 3 units. (b) What are the domain and range of f?
Answer: Domf = {x > 2 }, Ranf = (โโ, โ) (c) Find the inverse of f , or explain why it does not exist.
Answer: f โ^1 (x) = exโ^3 + 2
- For each of the following functions, state the domain and range.
(a) f (x) = ln x Answer: Domf = (0, โ), Ranf = (โโ, โ). (b) g(x) = (^) (x+3)^12 Answer: Domg = {x 6 = 3}, Rang = (โโ, โ3)โช(โ 3 , โ)
(c) The inverse of
3
)x Answer: Domain = (0, โ), Range = (โโ, โ).
- For each of the following, solve for x.
(a) log 2 x = 4 Answer: x = 2 (b) 3(x (^3) โ1) = b, where b > 0. Answer: x = 3
log 3 b + 1. (c) e^3 x^ = โ 1 Answer: No solution exists. (d) ln x + ln(x โ 3) = 2 ln 2 Answer: x = 4. Note that x = โ 1 is not a solution since -1 is not in the domain of ln x.
- A bacteria population doubles every 16 hours. Suppose that the initial population is
(a) How many bacteria are present after 32 hours? Answer: 400 (b) Find a formula for p(t), the size of the population after t hours. Answer: p(t) = 200(2t/^16 ). (c) At what time t are there 5000 bacteria present? (You do not need to simplify your answer!) Answer: 16 log 2 (20) or 16ln 20ln 2
- Let f be a function that satisfies all of the given conditions:
lim xโโ 2 f (x) = 1, f (โ2) = 3, lim xโ 0 f (x) = 0, f (0) = 0
lim xโ 1 โ^
f (x) = 0, lim xโ 1 +^
f (x) = โ 1.
(a) At each of the values, -2, 0, and 1, decide if f is continuous or discontinuous and state your reasons in terms of the limits given above.
-2: f is discontinuous because f (โ2) 6 = limxโโ 2 f (x) 0: f is continuous because f (0) = 0 = limxโ 0 f (x) 1: f is discontinuous because limxโ 1 f (x) is undefined.
(b) Sketch the graph of an example of a function satisfying the conditions given above.
There are plenty of graphs that will work here. It should have a hole at -2, be continuous at 0 and have a jump at 1.
(b) Let p(x) = x^4 + 7x โ 4. Is there a solution to p(x) = 0 for x in the interval (0, 1)? Justify your answer. Answer: Yes, since p(x) is a polynomial it is continuous everywhere, and in par- ticular it is continuous on the interval (0,1). We also notice that p(0) = โ 4 < 0 and p(1) = 4 > 0 so by the IVT there must be some x in the interval (0, 1) at which p(x) = 0.
- Find a formula for a function with vertical asymptotes at x = โ1 and x = 5 and horizontal asymptote at x = 3. Answer: f (x) = 3 x
2 (x+1)(xโ5)
- Recognize the limit as a derivative, f โฒ(a). Identify the function f and the value a :
lim xโ 8
โ (^3) x โ 2
x โ 8
Answer: This is the formula for the derivative of f (x) = 3
x at a = 8.
- Use the definition of the derivative to find the derivatives of the following functions at the indicated value a: (a) y = x^3 + 6x โ 2 ; a = 2 Answer: 18 (b) f (x) =
x + 3 ; a = 6 Answer: 1/
- Find the equation of the tangent line to the graph of y = x^2 + 2x at the point (1, 3).
Answer: y โ 3 = 4(x โ 1)
- For a graph of f be able to say, with reasons, at what points f is not differentiable, where the derivative is zero, positive and negative, and then draw the graph of the derivative. For example, any of problems 4-11 on page 162. See the solutions in the book.
