Recent content by swears

Homework Statement f: B => C and g: A => B 1. If f of g is injective, then f is injective. 2. If f of g is injective, then g is injective.Homework Equations The Attempt at a SolutionI know that 1 is true and 2 is false because I found those as properties, but I am not exactly sure why, and...

Well actually, f'(x) is only 0 at x= 0 and 2, but it switches concavity at 1, so I figured that would be one too. (guess not) x^4, yeah I believe so, I don't have a calculator to confirm(working on it), but it has two parabolas in it.(1 upside down).

It is going up until x=0, then goes down. It is concave down there. The at x = 1 it change to concave up and at x = 2 it starts increasing. So, I guess the critical points are x = 0, 1, 2

Yeah, I can do the derivative if I see the graph. I just can't visualize that backwards in my head. These are the values I'm dealing with: f'(x)| 0 | 1 | 0 | -1 | 0 ............x |-1 | 0 | 1 | 2 | 3

Wow, I totally forgot how to do these. I have a graph of a line, but it does not say the function. I have to sketch the antiderivative. Does anyone have any links or advice to start me out. TIA

bump for a test tomorrow

yeah, i think it's \frac{n (n+1)(2n+1)}{6} I'm not really sure I understand what you wrote above that though.

Ok, well i guess i can't do it at all. I have the answer, but no idea how to get there. I think the formula is \frac {b-a}{n} \sum^n_{i=1}F(Xi) So when I plug it in I get \frac {1-0}{n} \sum^n_{i=1}[2(x_{i})^2 +4] Is this right so far, and if so, what's next?

Use the Riemann sum and a limit to evaluate the exact area under the graph of y = 2x^2 + 4 on [0, 1] I know how to do this normally but now they ask to do it w/ a limit and I'm not sure how. (LaTex corrected by HallsofIvy.)

Thanks for the help d_leet.

Yeah, well my equation is not in your form. I don't have those x and y subzeroes.

Ah ok, So do I assume that the center of the circle is the origin here?

Ok Thanks. 2 more questions In the equation of a circle. What do the x subzero's stand for. And Now I figured out the area to be 25.13 but why do they give me from 0 to 4.

No. But how do you know the radius is 4? Are ou taking the difference from the x points? Wouldn't the radius be y axis.

From 0 to 4. ( I don't know how to make the s sign.) squareroot of (16-x^2) I set it equal to y and squared it to get. y^2 = 16 - x^2 Then I changed it to: Y^2 + X^2 = 16 I know it's a semi circle on the x-axis at 0 to 4. What should my final answer be?