Answer:
The absolute maximum and minimum of [tex]f(x) = 5 + 54\cdot x -2\cdot x^{3}[/tex] on [tex][0,4][/tex] are 113 and 5.
Step-by-step explanation:
Let be [tex]f(x) = 5 + 54\cdot x -2\cdot x^{3}[/tex], the first and second derivatives of the function are, respectively:
[tex]f'=54-6\cdot x^{2}[/tex]
[tex]f''=-12\cdot x[/tex]
Now, let equalize the first derivative to zero and solve the resulting expression:
[tex]54-6\cdot x^{2} = 0[/tex]
[tex]x^{2} = 9[/tex]
[tex]x =\pm 3[/tex]
According to the given interval, only [tex]x=3[/tex] is a valid outcome. Lastly, this is evaluated in the second derivative expression:
[tex]f''=-12\cdot(3)[/tex]
[tex]f'' = -36[/tex]
[tex]x = 3[/tex] leads to an absolute maximum.
[tex]f(3) = 5 + 54\cdot (3) -2\cdot (3)^{3}[/tex]
[tex]f(3) = 113[/tex]
The absolute minimum is determined by evaluating at each extreme of the interval:
[tex]x = 0[/tex]
[tex]f(0) = 5 + 54\cdot (0) -2\cdot (0)^{3}[/tex]
[tex]f(0) = 5[/tex]
[tex]x = 4[/tex]
[tex]f(4) = 5 + 54\cdot (4) -2\cdot (4)^{3}[/tex]
[tex]f(4) = 93[/tex]
The absolute maximum and minimum of [tex]f(x) = 5 + 54\cdot x -2\cdot x^{3}[/tex] on [tex][0,4][/tex] are 113 and 5.