Answer to Question #160741 in Calculus for sfda

Question

Locate the absolute maximum and minimum for each of the following functions and justify your responses. Be careful on your differentiation, especially with #2. Show all work and justify all answers

Question 1-

f(x)= (cube root of x^2)-x on the interval [-1,1]

Question 2-

𝑔(𝑥) = 𝑥𝑒^2𝑥 on the interval [−2,0] 

Answer 1

Solution

Question 1:

f’(x)= 2/(3*cube root of x)-1

Solve equation f’(x) = 0 => cube root of x = 2/3  =>  x₀ = 8/27.

Another point of f’(x) is x=0, where f’(0+) = ∞, f’(0-) = -∞. First derivative is not exist. But f(0) = 0.

f’’(x) = -2/(9*cube root of x^4)

For x>0 f’’(x)<0. Therefore x₀ = 8/27 is local maximum. f(x₀) = 0.148

It is necessary to check the value of the function at the boundary points of the interval [-1,1].

f(-1) = 2, f(1) = 0.

From calculated values: absolute maximum of f(x) is 2 for x = -1, absolute minimum is 0 for x = 0 and x = 1.

Question 2

g’(x) = e^2x+2xe^2x = (1+2x)e^2x

Solve equation g’(x) = 0  =>  x₀ = -1/2

g’’(x) = 2e^2x+2(1+2x)e^2x = 4(1+x)e^2x

For x = -1/2 f’’(x)>0  =>  x₀ is a point of local minimum. g(x₀) = -0.184

Let’s check the value of the function at the boundary points of the interval [--2,0].

g(-2) = -0.037, g(0) = 0.

From calculated values: absolute maximum of g(x) is 0 for x = 0, absolute minimum is -0.184 for x = -1/2.

Answer

1. Absolute maximum of f(x) is 2 for x = -1, absolute minimum is 0 for x = 0 and x = 1.
2. Absolute maximum of g(x) is 0 for x = 0, absolute minimum is -0.184 for x = -1/2.