(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
#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
Sure, I'd be happy to help you with these problems step-by-step! Here are the solutions to each problem:
1. If g(x) = 5^x, then g'(2) can be found by taking the derivative of g(x) with respect to x and evaluating it at x = 2. Taking the derivative, we have g'(x) = ln(5) * 5^x. Plugging in x = 2, we get g'(2) = ln(5) * 5^2 = ln(5) * 25. So the answer is 25ln(5).
2. To determine which of the following are equal to π/3, we can simplify each expression step-by-step:
- arctan(tan(π/3)): Since tan(π/3) = √3, arctan(√3) gives us π/3.
- arcos(sin(-π/6)): Since sin(-π/6) = -1/2, arcos(-1/2) gives us π/3.
- arcsin(sin((2π)/3)): Since sin((2π)/3) = √3/2, arcsin(√3/2) gives us π/3.
So all three expressions are equal to π/3.
3. To find the slope of the tangent line to the curve g^(-1)(x) at the point (2, 1), we first need to find the derivative of g^(-1)(x). Note that g^(-1)(x) is the inverse function of g(x). To find the derivative, we use the formula g'(x) = 1 / g^(-1)'(g(x)). Since g(x) = x^3 + 2x - 1, we have g^(-1)(x) = (x + 1)^(1/3). Taking the derivative of g^(-1)(x), we get g^(-1)'(x) = 1 / (3(x + 1)^(2/3)). Plugging in x = g(2) = 15, we have g^(-1)'(15) = 1 / (3(15 + 1)^(2/3)) = 1 / (3 * 16^(2/3)). Simplifying this expression gives us g^(-1)'(15) = 1 / 48. Therefore, the slope of the tangent line to the curve g^(-1)(x) at the point (2, 1) is 1 / 48.
4. To find the volume of the solid generated when the region enclosed by the system of equations is rotated about the x-axis, we need to use the method of cylindrical shells. The solid is enclosed by y = 0, x = 5, and y = 1/(sqrt(1 + x)). First, we need to find the limits of integration. The graph of y = 0 intersects the x-axis at x = 0. The graph of y = 1/(sqrt(1 + x)) intersects the x-axis at x = 3. Therefore, the limits of integration are from x = 0 to x = 3. Now, we need to find the radius and height of each cylindrical shell. The radius is given by r(x) = x, and the height is given by h(x) = 1/(sqrt(1 + x)). The volume of each cylindrical shell is given by V = 2πrh(x)Δx. Integrating V from x = 0 to x = 3, we obtain the volume of the solid. The result of the integration is π(ln(4) + 2ln(6) - 3ln(3)). Simplifying this expression gives us the volume of the solid as πln(8/9).
5. To evaluate the integral ∫(1)/(x^2 + 4)dx, we can use the trigonometric substitution method. Let x = 2tan(θ), which implies dx = 2sec^2(θ)dθ. Substituting these values into the integral, we have ∫(1)/(x^2 + 4)dx = ∫(1)/(4tan^2(θ) + 4)(2sec^2(θ))dθ. Simplifying this expression gives us ∫(1)/(8tan^2(θ) + 8)sec^2(θ)dθ. Rearranging the terms further gives us ∫(1/8)(1/cos^2(θ))dθ. Since 1/cos^2(θ) = sec^2(θ), this integral simplifies to ∫(1/8)sec^2(θ)dθ. The integral of sec^2(θ) is tan(θ). Therefore, the final result is (1/8)tan(θ) + C. Converting back to x, we have (1/8)tan(θ) + C = (1/8)tan(arctan(x/2)) + C = (1/8)(x/2) + C = (1/16)x + C. So the answer is (1/16)x + C.
I will provide the step-by-step solutions to the remaining problems in another response to avoid exceeding the character limit.
Sure! I can help you with these problems and explain how to solve them. Let's go through each question one by one:
1. To find g'(x), the derivative of g(x), we can use the power rule of differentiation. The derivative of 5^x is given by multiplying the base (5) by the natural logarithm of the base (ln(5)) and multiplying it by the original function: g'(x) = ln(5) * 5^x. To find g'(2), we substitute x = 2 into the derivative expression: g'(2) = ln(5) * 5^2 = 25ln(5). Therefore, the answer is 25ln(5).
2. To determine which expressions are equal to pi/3, we need to evaluate each expression.
- arctan(tan(pi/3)): The tangent of pi/3 is sqrt(3), and taking the inverse tangent of sqrt(3) gives pi/3.
- arccos(sin(-pi/6)): The sine of -pi/6 is -1/2, and taking the inverse cosine of -1/2 gives pi/3.
- arcsin(sin((2pi)/3)): The sine of (2pi)/3 is sqrt(3)/2, and taking the inverse sine of sqrt(3)/2 gives pi/3.
Therefore, all three expressions are equal to pi/3.
3. To find the slope of the tangent line to the curve g^-1(x) at the point (2,1), we need to find the derivative of g^-1(x) and evaluate it at x = 2. The inverse function g^-1(x) can be found by switching the roles of x and y in the equation g(x) = x^3 + 2x - 1 and solving for y. The resulting equation is y = (x - 2)^(1/3). Taking the derivative of g^-1(x) using the power rule gives (1/3)(x - 2)^(-2/3). Substituting x = 2 into the derivative expression gives (1/3)(2 - 2)^(-2/3) = (1/3)(0)^(-2/3) = undefined.
Therefore, the slope of the tangent line to the curve g^-1(x) at the point (2,1) is undefined.
4. To find the volume generated by rotating the region enclosed by the given system of equations about the x-axis, we need to use the method of cylindrical shells. The region is bounded by the x-axis (y = 0), the vertical line x = 5, and the curve y = 1/sqrt(1+x). The volume of a cylindrical shell with radius r and height h is given by the formula V = 2πrh.
To set up the integral, we need to express r and h in terms of x. Since the shells are perpendicular to the x-axis, h is equal to the height of the region, which is y = 1/sqrt(1+x). The radius r is equal to the distance from the axis of rotation (the x-axis) to the curve, which is y. Therefore, r = 1 / sqrt(1+x).
The integral for the volume is then given by V = ∫(2πrh)dx, with the limits of integration from 0 to 5.
After evaluating the integral, you will find that the volume is given by pi ln(6).
5. To evaluate the definite integral ∫(1)/(x^2 + 4)dx, we can use a substitution method. Let u = x^2 + 4, then du = 2xdx. Rearranging gives dx = du/(2x). Substituting these values into the integral expression gives ∫(1)/(x^2 + 4) * (du/(2x)).
Simplifying, we have (1/2)∫(1/u)du. Integrating (1/u) gives ln|u|. Substituting back into the expression, we get (1/2)ln|x^2 + 4| + C, where C is the constant of integration.
Therefore, the answer is 1/2 ln|x^2 + 4| + C.
I will continue with the explanations for the remaining questions in separate responses due to character limitations.