Answer
Absolute maximum at (3,113)
Absolute minimum at (0,5)
Work Step by Step
Original equation:$ f(x)=5+54x−2x^{3}$ on the interval $[0,4]$
To find the absolute maximum and minimum, we first need to find all the critical points of the function, as the local extrema might be the absolute extrema. Critical points of a function occur when a function's derivative is undefined or the derivative is zero.
In this case, we see that the function is a polynomial, which means it is defined on all intervals from $(−\infty,\infty)$. Therefore, we need to set the derivative to zero.
Finding the derivative of the function through the power rule:
$f′(x)=54-6x^{2}$
Then, we set this derivative to zero and solve for x using elementary algebra. We see that the critical point occurs when x=2
$54-6x^{2}=0$
$6x^{2}=54$
$x^{2}=9$
$x=3, -3$
In our given interval $[0,4]$, the only critical point we will consider is when $x=3$ as $x=-3$ is outside of our given interval.
Now we have three possible points of where a absolute minimum or maximum may occur: the critical point $x=3$, as well as the two endpoints of the interval, where $x=0,4$. To find which one of these is an absolute extrema vs a local extrema, we plug these x values back into the original equation and see what we get.
$f(0)=5+54(0)−2(0)^{3}=5$
$f(3)=5+54(3)−2(3)^{3}=113$
$f(4)=5+54(4)−2(4)^{3}=93$
Of the three values, we see that the minimum y value occurs when x=0 and the maximum y value occurs when x=3. Therefore, we have an absolute maximum at $(3,113)$ and an absolute minimum at $(0,5)$.
