- #1
frosty8688
- 126
- 0
1. Sketch the curve
2. [itex]\frac{x}{(x-1)^{2}}[/itex]
3. Here's what I have; the domain is all x except x ≠ 1, y-intercept of 0, x-intercept of 0, symmetric about the origin since it is an odd function, horizontal asymptote of 0, vertical asymptote of 1, and for the increasing and decreasing of the function, I'm wondering if I did it right. f'(x) = [itex]\frac{1*(x-1)^{2}-x*2(x-1)*1}{(x-1)^{4}}[/itex] = [itex]\frac{(x-1)^{2}-2x(x-1)}{(x-1)^{4}}[/itex] = after factoring out the one (x-1) [itex]\frac{-x-1}{(x-1)^{3}}[/itex] Increasing on (-1,1) and decreasing on (-∞,-1), (1,∞). There is no local maximum, because f(1) would result in division by 0 which is impossible. Local minimum of f(-1) = [itex]\frac{-1}{4}[/itex]
2. [itex]\frac{x}{(x-1)^{2}}[/itex]
3. Here's what I have; the domain is all x except x ≠ 1, y-intercept of 0, x-intercept of 0, symmetric about the origin since it is an odd function, horizontal asymptote of 0, vertical asymptote of 1, and for the increasing and decreasing of the function, I'm wondering if I did it right. f'(x) = [itex]\frac{1*(x-1)^{2}-x*2(x-1)*1}{(x-1)^{4}}[/itex] = [itex]\frac{(x-1)^{2}-2x(x-1)}{(x-1)^{4}}[/itex] = after factoring out the one (x-1) [itex]\frac{-x-1}{(x-1)^{3}}[/itex] Increasing on (-1,1) and decreasing on (-∞,-1), (1,∞). There is no local maximum, because f(1) would result in division by 0 which is impossible. Local minimum of f(-1) = [itex]\frac{-1}{4}[/itex]