Find the absolute maximum and minimum values of the function f (x) = [x^(2/3)]*[x − 1] on the
closed interval [0, 4], and state where these values occur.
f' (x) = x^(2/3) + (2/3)x^(-1/3) (x-1)
= (1/3)x^(-1/3) [ 3x - 2(x-1) ]
= (1/3)x^(-1/3) [ x + 2]
= 0 for a max/min
x = 0 or x = -2
checking the f(x) for those and the endvalues
f(0) = 0
f(-2) = (-2)^(2/3) (-2-1) = -3(-2)^(2/3)
= appr 4.76
f(4) = 4^(2/3) (4-1) = appr 7.56
so the max is 3( 4^(2/3) ) or appr 7.56
and the min is 0
thank you very much
To find the absolute maximum and minimum values of the function f(x) = x^(2/3) * (x - 1) on the closed interval [0, 4], we can follow these steps:
1. Determine the critical points:
- Take the derivative of the function, f'(x), using the product rule.
- Set f'(x) equal to zero and solve for x to find the critical points.
- In this case, we only have one critical point, which is x = 1.
2. Evaluate the function at the critical points and endpoints:
- Plug in the critical point, x = 1, and the endpoints, x = 0 and x = 4, into the original function f(x).
- Calculate the corresponding function values at each of these points.
3. Compare the function values:
- The largest function value will be the absolute maximum, and the smallest function value will be the absolute minimum.
- Determine where the maximum and minimum occur by identifying the corresponding x-values.
Let's go through these steps to find the absolute maximum and minimum values of the given function f(x) = x^(2/3) * (x - 1) on the closed interval [0, 4]:
1. Determine the critical points:
- Taking the derivative of f(x) = x^(2/3) * (x - 1), we get:
f'(x) = (2/3)x^(-1/3) * (x - 1) + x^(2/3) * 1
Simplifying f'(x), we get:
f'(x) = (2/3)x^(-1/3) * (x - 1) + (3/3)x^(2/3)
f'(x) = (2/3)x^(-1/3) * (x - 1) + (3/3)x^(2/3)
f'(x) = (2/3)x^(-1/3)(x - 1) + (3/3)x^(2/3)
f'(x) = (2/3)(x^(-1/3))(x - 1) + (3/3)(x^(2/3))
- To find the critical point, set f'(x) = 0:
(2/3)(x^(-1/3))(x - 1) + (3/3)(x^(2/3)) = 0
Multiply through by 3x^3 to clear the fractions:
2(x - 1)x^(2/3) + 3(x^2) = 0
2x(x - 1)x^(2/3) + 3x^2 = 0
2x^2 - 2x + 3x^2 = 0
5x^2 - 2x = 0
x(5x - 2) = 0
- We get two potential critical points: x = 0 and x = 2/5 (which is 0.4).
2. Evaluate the function at the critical points and endpoints:
- Plugging these x-values into the original function f(x), we get:
f(0) = (0^(2/3))(0 - 1) = 0(0 - 1) = 0
f(2/5) = ((2/5)^(2/3))((2/5) - 1) = (4/25)(2/5 - 1) = (4/25)(2/5 - 5/5) = (4/25)(-3/5) = -12/125
f(4) = (4^(2/3))(4 - 1) = (8/4)(3) = 2(3) = 6
3. Compare the function values:
- The function values are: f(0) = 0, f(2/5) = -12/125, and f(4) = 6.
- The absolute maximum value is 6, which occurs at x = 4.
- The absolute minimum value is -12/125, which occurs at x = 2/5 (or 0.4).
Therefore, the absolute maximum value of the function f(x) = x^(2/3) * (x - 1) on the closed interval [0, 4] is 6, occurring at x = 4. The absolute minimum value is -12/125, occurring at x = 2/5 (or 0.4).