A few questions - also would like someone to check my work

[SniXSniPe]

The one's I have absolutely no idea how to do:
9. f(x) = x^3/2 - 1, find the solution to the equation f(x) = f^-1(x)

2. If the solution to the polynomial inequality p + x - 2x^2 > 4 is -1 < x < 3/2, which of the folllowing could be the value of p?

A. 0
B. 1
C. 3
D. 4
E. 7

I'm not just interested in the answers--- but on how to SOLVE the problems.

And... Just some I want to make sure are correct.

5. Which of the folllowing gives all of the values of x for which the conic section x^2 + y^2 + 6x - 4y = 12 is defined?

My answer: [-8, 2

2. What is the area of the region defined by the inequality x^2 + y^2 + 4x - 6y < 3?

My answer: 16pi

Thanks for looking at this.

One last thing: Does anyone know of a site that can help teach me in the following pre-calculus topics: Exponential & Logarithmic Functions, and Functions

I'm terrible at functions. I don't want to post all the other questions I don't know how to do, because I don't want other people to do my work.

 

[stapel]

9) Does the question really ask you to solve "f(x) = f^-1(x)"? Or does it perhaps ask you to find f^-1(x), given f(x)?

2) The related equality, p + x - 2x² = 4, may be rearranged as 0 = 2x² - x + (4 - p). Plug this into the Quadratic Formula. Set the solutions (in terms of p) equal to -1 and 3/2. Solve for p.

5) How did you arrive at these values? You completed the square to find the equation of the circle, read off the center and the radius, and... then what?

2) Since the radius r = 4, then the area A is 16(pi).

There are many sites that cover various aspects of functions, exponentials, and logarithms. But these topics are fairly broad. You might want to try doing a search (such as with Google) for keywords from the section or chapter in question. Or else please narrow the request. For instance, are you needing to graph logarithmic functions in non-calculator-friendly bases? Or something else?

Thank you.

Eliz.

 

[tkhunny]

Re: A few questions - also would like someone to check my wo

The one's I have absolutely no idea how to do:
9. f(x) = x^3/2 - 1, find the solution to the equation f(x) = f^-1(x)

Something wrong with this one. If \(\displaystyle f(x) = f^{-1}(x)\), then f(x) = x. What has that to do with where you started?

2. If the solution to the polynomial inequality p + x - 2x^2 > 4 is -1 < x < 3/2, which of the folllowing could be the value of p?

Wouldn't \(\displaystyle \L\,-2(-1)^{2} + (-1) + p = 4\) solve this one?

5. Which of the folllowing gives all of the values of x for which the conic section x^2 + y^2 + 6x - 4y = 12 is defined?

Missing a square bracket there on the right?

2. What is the area of the region defined by the inequality x^2 + y^2 + 4x - 6y < 3?

Lovely answers. Too bad we've no clue how you found them.

 

[SniXSniPe]

Err--- I see I didn't list the answers for this one:

#9. If F(x)=x^(3/2) - 1, find the solution to the equation f(x) = f^-1(x)
A. 1
B. 2.148
C. 1.249
D. 1.587
E. There is no solution.

That's exactly how it's stated.

In my packet, this is classified as "Pre-Calculus Unit 1 (Calculator): Functions". However, I am taking Pre-Cal next year... this is the summer packet--- unfortunately there is no information on how to do most of these :/.

#2. If the solution to the polynomial inequality p + x - 2x^2 > 4 is -1 < x < 3/2, which of the folllowing could be the value of p?

I got 4 as an answer by subtracting 1 from 1.5 to get 0.5, and then substituting 0.5 for x. The equation is then p + 0.5 - 2(0.5)^2 = 4

So you are then left with p = 4, since 0.5 and -2(0.5)^2 cancel each other out.

---Is that wrong?

5. Which of the folllowing gives all of the values of x for which the conic section x^2 + y^2 + 6x - 4y = 12 is defined?

My answer: [-8, 2]

Equation: (x + 3)^2 + (y - 2)^2 = 25
Center: (-3, 2)
Radius: r = 5
So the x-values should be from [-8,2) since it's a circle

 

[stapel]

9) I will assume that you mean "F(x)" to be the same as "f(x)". This is not standard mathematical practice, so please pardon me if my assumption is incorrect.

As was noted earlier, the function f(x) equals its inverse, f^-1(x), only for the function f(x) = x. Since this is not the function you were given, the question, as written, makes no sense. So I will guess that the question actually means to ask "find the value of x for which f(x) equals f^-1(x)".

In this case, you need first to find the inverse function. Once you have the expression for f^-1(x), set this equal to x^3/2 - 1, and solve the resulting equation for x.

2) I'm sorry, but I don't follow your reasoning.

5) This looks good to me.

Eliz.

 

[tkhunny]

I see. I misread the first one. You are looking for a stationary point - a point that is mapped to itself. Solve this: x^(3/2) - 1 = x -- Then tell me why that worked.

On the last one, why is it missing x = 2? It wasn't a moment ago, but managed to lost it along the way.

 

[skeeter]

... the function f(x) equals its inverse, f^-1(x), only for the function f(x) = x.

not so fast ... what about the function f(x) = 1/x ?

 

[tkhunny]

Thanks for blaming that on stapel. Too bad I said it, too.

SniXSniPe, do you know how to find an inverse ralation?

 

[pka]

Re: A few questions - also would like someone to check my wo

The one's I have absolutely no idea how to do:
9. f(x) = x^3/2 - 1, find the solution to the equation f(x) = f^-1(x)

I think that the above question means the following.
\(\displaystyle \L
f(x) = x^{\frac{3}{2}} - 1\quad \Rightarrow \quad f^{ - 1} = \left( {x + 1} \right)^{\frac{2}{3}}\)

Thus we asked to solve this equation:
\(\displaystyle \L
x^{\frac{3}{2}} - 1 = \left( {x + 1} \right)^{\frac{2}{3}} \quad \Rightarrow \quad x = 2.148.\)