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Calculus (Help please!)

(Re-posting because I think the first one got lost in all the questions)
This isn't a test or assignment, it's a review for studying. I'm trying to catch up in my lesson as I have fallen quite behind due to health issues. I'm here for help because I have no understanding of these problems. I was hoping someone could explain them and give the answers so I could use these to study similar problems. (If you could solve it completely it would be preferred so then I can get back to you on the individual steps that confuse me) Thank you so much to anyone who is taking the time to read this and help me out!

1.
If g(x)=5^x, then g'(2) is:
5
25In5
1/In5
5/In5
5 In5

2.
Which of the following are equal to pi/3?
arctan(tan pi/3)
arcos(sin -pi/6)
arcsin(sin ((2pi)/3)

3.
g(x)=x^3+2x-1 What is the slope of the tangent line to the curve g^-1 (x) at the pont (2,1)?
5
1/5
1
14
8

4.
The region in the first quadrant enclosed by the system of equations is rotated about the x-axis. The volume of the solid generated is:
y=0, x=5, and y=1/(sqrt(1+x))
pi In(6)
pi In(5)
pi In(4/3)
pi In(3/4)
pi In(10)

5.
∫(1)/(x^2 +4)dx=
arctan(2x)+C
1/2 arctan((1/2)x)+C
1/2 arctan(x)+C
In|x^2 +4| +C
1/2 In|x^2 +4|+C

6.
lim ((x-3)/(log3(x-1))
x->2
2
log3^2
-1/log3^2
infinity
does not exist

7.
The graph of f(x) is shown along with the line y=x
Assume that f^-1 (x) is defined by restricting f(x) to the domain a<x<b. Which of the following statements is false?
Graph (gyazo.com/c6d4871b661f5ed9207e169542ac4a79)
f^-1 (x) is a one to one function
f^-1 (x) has a domain of d<x<c
a=f^-1 (f(a))
(d/dx)(f^-1(x)) >=0
f^-1(c)=a

8.
The slope of the line normal to the curve e^x-x^3+y^2=10 at the point (0,3) is:
-1/6
1/6
6
-6
Undefined

9.
x
∫ t/(1+t^4) dt =
0
1/2arctan(x^2)
1/2arctan(x^4)
1/2In(1+x^4)
-1/2In(1+x^4)
1/2arcsin(x^2)

10.
Given (table: gyazo.com/daa1e2fb3a5c1b0bbe9afdccae6b56d7) what is a reasonable value for (d/dx)h^-1(x) evaluated at x=5?
3
1/3
1
-3
-1/3

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  1. #1
    g(x) = 5^x
    g'(x) = ln5 * 5^x
    g'(2) = ln5 * 5^2 = 25ln5

    #2
    tan(x) and arctan(x) are inverse functions.
    So, by definition, arctan(tan(x)) = tan(arctan(x)) = x
    not technically always true, but it holds if we restrict arctan(x) to its principal values.

    #3
    Let f(x) = g^-1(x) then
    if g(a) = b then f'(b) = 1/g'(a)
    g'(x) = 3x^2+2
    so, you have f(2) = 1, so g(1) = 2
    That means f'(2) = 1/g'(1) = 1/5

    #4
    v = ∫[0,5] π*1/(1+x) dx = π(ln6 - ln1) = π ln6

    #5
    this just a standard form: ∫ dx/(x^2+a^2) = 1/a arctan(x/a)

    #6
    the limit is (2-3)/log(2-1) = -1/0 DNE
    the base of the log does not matter. log(1) = 0 always

    #7
    f^-1 (x) is a one to one function -- true since f passes the vertical-line test
    f^-1 (x) has a domain of d<x<c -- true since f has a range of d to c
    a=f^-1 (f(a)) -- false, since f(a) is not defined (the domain does not include a)
    (d/dx)(f^-1(x)) >=0 -- hmmm. also false -- see #3
    f^-1(c)=a -- true, but not if we restrict the domain, since f(a) is not defined.

    #8
    e^x-x^3+y^2 = 10
    e^x - 3x^2 + 2yy' = 0
    y' = (3x^2 - e^x)/(2y)
    at (0,3) y' = -1/6
    so the normal has slope 6

    #9
    if u = t^2 then we have du = 2t dt
    ∫[0,x] t/(1+t^4) dt = ∫[0,x^2] 1/2 du/(1+u^2)
    = 1/2 arctan(u) [0,x^2] = 1/2 arctan(x^2)

    #10
    h(1) = 5
    h'(1) ≈ -.3/.1 = -3
    so, [h^-1(5)]' = 1/h'(1) ≈ -1/3

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    oobleck

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