# Y= 8e^7x+8e^-7x (simplify Answer,type A Ordered...

 AnonymousFrom: -Posts: -Votes: - Find the derivative: y ' = 56e7x - 56e-7x Set the derivative equal to zero: 0 = 56e7x - 56e-7x Observe that neither term can ever equal zero...but their difference can equal zero only when x iszero. Thus, when x is zero, y' is zero. Evaluate f (0): f (0) = 8e0 + 8e0 = 16 Observe that the sign of the first derivative to the left ofzero is negative, indicating a decreasing function, and the sign ofthe first derivative to the right of zero is positive, indicatingan increasing function. Thus, there is a minimum at x = 0, y =16 (or at (0, 16)). Thus, (0, 16) is youranswer. Mike

 AnonymousFrom: -Posts: -Votes: - Y'=56e7x-56e-7x=0 56(e7x-e-7x)=0 e7x=e-7x lne7x=lne-7x 7x=-7x x=0 is the only critical point. --> y(0)=8e^0+8e^0=8+8=16. y'(-1)=56e-49-56e49<0 (decreasing) y'(1)=56e49-56e-49>0 (increasing) y goes from decreasing to increasing, which means that (0,16) is arelative minimum. There is no maximum, as the function increase toward ∞ to theleft or right.

 LovelySupa3From: USPosts: 54Votes: 162 Y =8e^7x+8e^-7x take the derivative y = 8e7x(7) +8e-7x(-7) simplify y' = 56e7x -56e-7x factor y' =56e-7x(e14x -1) find critcal number x =0 plugin x in the oringal equation y(0) = 8 + 8 = 16 point: (0, 16) <---- minimum take the second derivative to checkwhether the point is a minimum or maximum y' = 56e7x -56e-7x take the derivative y'' = 392e7x +392e-7x plug in 0 for y y''(0) = 784