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I'm trying to work through the review/practice euqations my professor recommened us to do for our test tomorrow, and I have a few problems that I have no clue what to do on and I'm not entirely sure what to google to figure them out. I have the answers for all of them, but no clue how to get them.


The one I'm trying to figure out right now is this:
EDIT:new issue posted

Assuming you know how to factor general quadratics, you can turn this kind of quartic polynomial into a quadratic by letting y=x^2. Then, you're dealing with a quadratic, since (x^2)^2 - 3(x^2) + 2 = y^2 - 3y + 2. So factoring that gives you (y-1)(y-2), which isn't exactly what you want quite yet. So you just substitute your original variables back in and you have (x^2 - 1)(x^2 - 2) and continue solving.

Ok, thanks, I think I've got that one now. This is the next one I'm stuck on:

2x(4-x)^-1/2 - 3sqrt(4-x) = 0

the answer for that one is 12/5, and I have on clue where to even begin this one or where I'm trying to get before the answer.

Nevermind, I believe I've got this one as well. Converted sqrt(4-x) to (4-x)^1/2 and eventually got:2x(4-x)^-1/2 = 3(4-x)^1/2. Divided both sides by (4-x)^-1/2, which gave me 2x = 3(4-x), from there it's a simple problem

Here's what I got.

x^4-3x^2+2 = 0
What wolf did was turn -3x^2 into -2x^2 -x^2. It still works out into -3x^2 but in this form, it more workable

x^4-2x^2 -x^2+2=0
Next was factor x^4-2x^2
x^2(x^2-2)-x^2+2=0
Next cross out x^2 and -x^2 and your left with
(x^2-2)-1 + 2 = 0
next add 1 and subtract 2 from both sides
(x^2-2)-1 + 2 + 1 - 2 = 0 + 1 - 2
You should get this
x^2 - 2 = -1
add 2 to both sides
x^2 = 1
so
x = 1

At least that's what I got. It's been a while since I've done this.

After finding the x, your problem, after replacing x with 1, you get 0 as the answer.

Ok, I think I did this one right but I'm not sure. I'm supposed to evaluate f(x+h)-f(x)/h when f(x) = x^2-5x-3.

This should equal 1, right? The only part I did that I'm not sure of is the f(x+h). f(x+h) = x^2-5x-3+h, right?

Edit: I have to find the domain of x/4sin^2(x)-1. So far, I've figured out that 4sin^2(x) > 1. But I'm not sure where to go from there.

The question dealing with (f(x+h)-f(x))/h is a lot more involved because it involves limits (and I believe your question should be asking you to evaluate the limit of that expression as h goes to 0). Briefly, f(x+h) is (x+h)^2 - 5(x+h) - 3 and the answer should be 2x-5.

Ok, another question (I've a feeling this thread is going to see a lot of use this semester).

Here's the problem from my h/w (my answers are bolded in brackets):

3. The point P (2,1/2) lies on the curve y=1/x.
a. If Q is the point (x,1/x), use your calculator to find the slope of the secant line PQ (correct to 6 decimal places) for the following values of x.

(i) 1.5 [-.333333] (ii) 1.9 [-.263158] (iii) 1.99 [-.251256] (iv) 2.5 [-.2] (v) 2.1 [-.238095] (vi) 2.01 [-.248756]

b. Using the results in part (a), guess the value of the slope of the tangent line to the curve at P (2, 1/2). [-.25]

c. Use algebra to show why this is a reasonable guess.

C. is what I'm stuck on. I have no clue how to show that my guess is reasonable. Anyone have an idea of what he's looking for for that part?

SnoringFrog (post: 1453705) wrote:Ok, another question (I've a feeling this thread is going to see a lot of use this semester).

Here's the problem from my h/w (my answers are bolded in brackets):

3. The point P (2,1/2) lies on the curve y=1/x.
a. If Q is the point (x,1/x), use your calculator to find the slope of the secant line PQ (correct to 6 decimal places) for the following values of x.

(i) 1.5 [-.333333] (ii) 1.9 [-.263158] (iii) 1.99 [-.251256] (iv) 2.5 [-.2] (v) 2.1 [-.238095] (vi) 2.01 [-.248756]

b. Using the results in part (a), guess the value of the slope of the tangent line to the curve at P (2, 1/2). [-.25]

c. Use algebra to show why this is a reasonable guess.

C. is what I'm stuck on. I have no clue how to show that my guess is reasonable. Anyone have an idea of what he's looking for for that part?

I imagine he is trying to get you to notice that the slope of the secant line approaches the slope of the tangent as the two points chosen become closer and closer to the point P. Since the derivative (slope) of this curve is -1/(x*x), then the slope at P is -1/4 or -0.25. Notice that the points in the previous part show a slope closer to -0.25 when the two points are nearer to P. By algebra, I assume he means using the 2-point solution to slope where:

slope = (y2-y1)/(x2-x1)

Hope this helps.

SP1 (post: 1454530) wrote:I imagine he is trying to get you to notice that the slope of the secant line approaches the slope of the tangent as the two points chosen become closer and closer to the point P. Since the derivative (slope) of this curve is -1/(x*x), then the slope at P is -1/4 or -0.25. Notice that the points in the previous part show a slope closer to -0.25 when the two points are nearer to P. By algebra, I assume he means using the 2-point solution to slope where:

slope = (y2-y1)/(x2-x1)

Hope this helps.

Ok, thanks. I think I understand that one now, sorta.

Now I'm stuck on a problem involving continuity that I think should be fairly easy, but I'm not sure.

If f and g are continuous functions with f(2)=4 and, find g(2).

Shouldn't the answer just be 4? If they are continuous with it, they would be the same, correct? But plugging in 4 for both values doesn't work out with that equasion.

Edit: Would it be .875? If f(x) and g(x) both multiply by 2 (which I guessed that they do since f(2) = 4 and they are continuous with f(2)=4), .875*2 = 1.75, and when you plug that into the equasion you get 7...but for some reason I think I did something wrong there. The other answer I came up with was g(2) = -5. Which is completely different.

Edit2: Also, I have no clue where to even start with these. I don't really want an answer, because thusfar I haven't even been able to make an attempt at them, but I don't even know how to approach these.

[quote]
3. Find all constants c that make f continuous on (-∞]

Edit3: Ok, I think I got number 3. The answers are 1 and -5. But I figured those out through trial and error, is there a better way to do that?

Sorry, I don't get on here everyday.

I think I missed something about the continuous function questions. Like, part of the problem is missing. To say that f and g are continuous functions and f(2)=4 has no bearing whatsoever on g(2), unless you know something more about these functions.

Whoops. Thought the rest of that got pasted in. If f and g are continuous functions with f(2)=4 and lim┬(x→2)⁡[3f(x)+g(x)]=7, find g(2).

I already took my guess and turned that homework in, but I'd still like to know if I was right, and my professor doesn't seem to give homework back at any point so I have no clue how I've done on any of it.

This has me stumped tonight:
If (x)=(x^2-3)/(x+1) , find f' (a) and use it to find the slope of the curve y=(x^2-3)/(x+1) at the points (1,-1) and (3,3/2).

I'm about to just give up on this one. I need more sleep because I've been sick lately anyways. Taking random guesses probably won't get me anywhere with this one. I know to take the derivative of (x^2-3)/(x+1) but other than that I'm not sure what to do.

Edit: the answers I came up with for the problem from tonight were these:
slope at (1,-1) = 2, and slope at (3, 3/2) = 9/8.

The value I used for f`(a) (after reorganizing/simplifying it plenty, so there could easily have been an error in there somewhere) was (a^2+2a+3)/(a^2+2a+1)

SnoringFrog (post: 1456648) wrote:Whoops. Thought the rest of that got pasted in. If f and g are continuous functions with f(2)=4 and lim┬]=7, find g(2).

Using the property of limits, you can separate that limit into a sum of limits. Since f and g are continuous, you know that lim f(x) = f(a), so you can replace lim f(x) with f(a) and lim g(x) with g(a) and solve for g(a).

SnoringFrog (post: 1456648) wrote:If (x)=(x^2-3)/(x+1) , find f' (a) and use it to find the slope of the curve y=(x^2-3)/(x+1) at the points (1,-1) and (3,3/2).

The derivative f'(a) of f(a) is defined to be the slope of the tangent at the point (a, f(a)).

Thanks for all of the help! It's been a huge...well, help XD

Thankfully the stuff we've been covering lately has been making more sense to me. I think I'm finally over that initial "holy crap what is this?!" curve.

Ok, two more issues now, but I think they're mostly the same. They're both related rates problems, which thusfar I've been able to get without much difficulty, but these two have me pretty lost.


1) A streetlight is at the top of a 20ft pole. A 5ft woman is walking away from the base of the pole at 4ft/sec. How fast is the tip of her shadow moving when she is 30ft from the pole?

I have a diagram drawn showing the light, woman and her shadow as a right triangle inscribed in a right larger triangle. The height is 20, the base is x+y (x being the distance from the pole to the woman, and y being the distance from the woman to the tip of her shadow). The height of the smaller triangle (which is formed by the woman) is 5.

I also have this proportion written down that the professor told us we needed: 20/x+y = 5/y. From what he said, I'm looking for dy/dt + dx/dt.

I tried working through it, and what I came out with was 16/3 ft/sec, but I have a feeling I did it wrong because I never had a place to plug the 30ft in, which I think I should have.




2) A spotlight is on the ground shining on a wall 16 meters away. A man who is 2m tall walks from the spotlight towards the wall at 1.6 m/sec. How fast is the length of his shadow decreasing when he is 5m from the building?

I've got a diagram drawn for this one too. The smaller triangle has a base of y and a height of 2. The larger triangle has a base of x+y (which is 16) and a height of a.

I tried solving this one starting from about the same spot I started the last problem from, but I ended up with dy/dt = 32/(da/dt), with no way to figure out da/dt. Again, I'm pretty sure I did something wrong here, but have no clue what. None of the examples we did involved these types of equations/diagrams, so I'm not sure what I'm missing here.