Recent content by nobelsmoke

Great Idea, I'll post here after I submit my final attempt.

Ok, thank you. I will keep that in mind.

Thank you for responding RedBelly98 Here's my work: f'(x) = 32e^(32x) f''(x) = 1024e^(32x) k(x) = (1024e^(32x)) / [1+1024e^(64x)]^3/2 k'(x) = (32768e^(32x) / (1024e^(64x)+1)^(3/2) - (100663296e^(96x)) / (1+1024e^(64x))^(5/2) when k'(x) = 0 , x = - log(2048)/64 y = e^(32x) y =...

What point did you come to? Here is a link to my two incorrect attempts: http://i.imgur.com/DDKEO.png

Firstly, HallsofIvy Thank you for responding. The formula for k(x) you have written is different than the one I quoted. In mine, 1 is inside the parenthesis in the denominator, also raised to the power of (3/2). Anyway, we know that abs(f''(x)) belongs in the numerator. And...

Here is the question: At what point does the curve y=e^(32x) have maximum curvature? I have tried this method: http://www.math.washington.edu/~conroy/m126-general/exams/mt2SolMath126Win2006.pdf Problem 4. Adapting for 32x rather than x it seems to get a bit lengthier with 32x than...