Originally Posted by

**DarkestEvil** 6.b. Consider the curve defined by the equation: $\displaystyle y+\cos y=x+1$ for $\displaystyle 0<y<2\pi$

Write an equation for each vertical tangent to the curve. (Hint: Vertical lines are x=#)

-I already found that $\displaystyle y'=\frac{1}{1-\sin y}$, but im not sure how to solve this one. how am i supposed to start off

vertical tangent lines occur where the derivative is undefined ...

1 - sin y = 0 ... solve for y

10. The expression $\displaystyle \frac{1}{4}(\sqrt{1}+2\sqrt{\frac{5}{4}}+2\sqrt{2} +2\sqrt{\frac{13}{4}}+\sqrt{5})$ is the trapezoid approximation **for which of the following definite integrals?**

This kind of expression always baffles me, it would really help if someone would care to explain

it might help to see the "integrals"

2.a. Given that f is the function defined by $\displaystyle f(x)=\frac{x^3-x}{x^3-4x}$

Find limit goes to infinity of f(x)

-limits have always confused me. **what am i supposed to be finding?**

the horizontal asymptote ... divide every term by the highest power of x and then take the limit as x goes to infinity

2.b. Write an equation for each vertical and each horizontal asymptote to the graph of f.

-i found that the vertical asymptotes are at x=0, x=2, x=-2 and horizontal at y=1, at least thats what i think. what am i supposed to with these?

this part is correct except x = 0 is not a vertical asymptote ... it is a removable discontinuity (a hole)