Answer :
The absolute maximum point for the function [tex]\(f(x) = 2x^3 - 4x^2 + 2\)[/tex]on the interval [tex]\(\left[\frac{1}{2}, 1\right]\) is \((\frac{4}{3}, \frac{22}{9})\).[/tex]
To find the absolute maximum point of the function [tex]\( f(x) = 2x^3 - 4x^2 + 2 \)[/tex]on the interval [tex]\( \left[\frac{1}{2}, 1\right] \),[/tex] we need to analyze the critical points and endpoints within this interval.
First, let's find the critical points by taking the derivative of f(x) and setting it equal to zero:
[tex]\[ f'(x) = 6x^2 - 8x = 0 \][/tex]
[tex]\[ 2x(3x - 4) = 0 \][/tex]
This gives us two critical points: x = 0 and [tex]\( x = \frac{4}{3} \)[/tex]. However, x = 0 is not within the interval [tex]\( \left[\frac{1}{2}, 1\right] \),[/tex] so we only need to consider [tex]\( x = \frac{4}{3} \).[/tex]
Now, let's find the values of f(x) at the critical point and the endpoints of the interval:
[tex]\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 4\left(\frac{1}{2}\right)^2 + 2 = \frac{1}{2} - 1 + 2 = \frac{5}{2} \][/tex]
[tex]\[ f\left(1\right) = 2(1)^3 - 4(1)^2 + 2 = 2 - 4 + 2 = 0 \][/tex]
[tex]\[ f\left(\frac{4}{3}\right) = 2\left(\frac{4}{3}\right)^3 - 4\left(\frac{4}{3}\right)^2 + 2 = \frac{32}{3} - \frac{64}{9} + 2 = \frac{22}{9} \][/tex]
Comparing these values, we see that [tex]\( f\left(\frac{4}{3}\right) = \frac{22}{9} \)[/tex] is the largest, so the absolute maximum point is [tex]\( \left(\frac{4}{3}, \frac{22}{9}\right) \).[/tex]
Therefore, the correct answer is option d. [tex]\( \left(\frac{4}{3}, \frac{22}{9}\right) \).[/tex]
The Correct question is:
Given the function f left parenthesis x right parenthesis equals 2 x cubed minus 4 x squared plus 2; locate the absolute maximum point for this function on the interval open square brackets 1 half comma space 1 close square brackets. a. (1, 0) b. open parentheses 1 half comma space 5 over 4 close parentheses c. only (0, 2) d. open parentheses 2 over 3 comma space 22 over 27 close parentheses
