MATH 165 Midterm: MATH 165 Iowa State MidtermPDMF16Fall2016

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turquoisegnu128

15 Feb 2019

School

Iowa State University

Department

Mathematics

Course

MATH 165

Professor

All

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Disclaimer: this is not the departmental midterm, and was initially written without reference to the departmental midterm. This may or may not have problems similar to the actual midterm. That being said, if you can do well on this practice midterm you also will do well on the departmental exam. Round 1: find the following limits: (a) lim x 1. 3x 1 x2 + 2x 3 x+2 (b) lim. 3 + 1 1 (c) lim t . 1 (c) find lim x 1 f(x) does not exist. or explain why it (d) find lim x 3 f(cid:0)f(x)(cid:1) or explain why it does not exist. Round 2: use the de nition of the derivative to nd the derivative of f(t) = (no credit for nding the derivative using other methods!) 4 t: find the horizontal asymptote(s) for f(x) = The following were originally problems that were included in a rst draft and draw from materials in section 3. 8.

Related Questions

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

1. Use the definition of continuity and the properties of limitsto show that the function is continuous at the given number(a).

f(x) = (x+3x^5)^4 , a= (-1)

lim f(x) as x approaches -1 (x+3x^5)^4

= (lim x +3 times lim x^5)^4

= (-1 +3 (???)^5)^4 ------got stuck here

= ???

f(-1) = ??

2. Find the average value f(ave) of the function f on the giveninterval

f(x)= 6x -x^2 , [0, 4]

3. Find the area of the shaded region between a curve and aline:

y = 9x-x^2 and y = 2x

the functions intersect at (0,0) and (7, 14)

I need to figure out the area of the upper half, above the y =2xline

it is a concave down parabola

MATH 165 Midterm: MATH 165 Iowa State MidtermPDMF16Fall2016

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