Calculus Question

Question

Find the points at which y=f(x)=x9−2x has a global maximum and minimum on the interval 0≤x≤2.4. Round your answers to two decimal places. 

Global Maximum: 

(x,y) = (_,_)

 Global Minimum: 

(x,y) = (_,_)

 

Answer 1

Global max/min is the say as a local max/min but need to look at the endpoints as well.

f(x) = x⁹ - 2x

f'(x) = 9x⁸ - 2 = 0 (find the critical points)

x⁸ = 2/9

x = (2/9)^1/8 ~ +/- 0.8286 (we can throw away the negative answer since it is outside our bounds)

f"(x) --> tells us about concavity: if f"(x) < 0 (max), if f"(x) > 0 (min)

f"(x) = 72x⁷

f"(x=0.8286) = positive number (i.e. minimum)

f(x=0.8286) ~ -1.473

Therefore let's now check the endpoints:

f(x=0) = 0

f(x=2.4) = 2637.008

So the global maximum would be f(x=2.4) --> (2.4, 2637.008)

The global minimum would be the same as the local minimum f(0.8286) ~ -1.473 --> (0.8286,-1.473)