y = x^11 - 7x
y' = 11x^10 - 7
so local extrema at x = ±(7/11)^(1/10) ≈ ±0.96
Our domain is [0,2.5] so we just have to worry about _0.96
Since y" > 0 there, that is a local minimum.
y(0) = 0
y(2.5) > 0, so
y(0.96) is the global min
y(2.5) is the global max
Find the points at which y=f(x)=x11−7x has a global maximum and minimum on the interval 0≤x≤2.5. Round your answers to two decimal places.
Global max:
Global min:
1 answer
1 answer