Find the absolute extrema of the function on the closed interval.f(x) = 3 − x, [−1, 2]minimum (x, y) = maximum (x, y) =
![Find the absolute extrema of the function on the closed interval.f(x) = 3 − x, [−1, 2]minimum (x, y) = maximum (x, y) = class=](https://us-static.z-dn.net/files/dea/003e0b5006993a80992282ce8ad25ff2.png)
Given:
There are given the function:
[tex]f(x)=3-x[/tex]Explanation:
To find the maxima and minima extrema, first, we need to find the difference between the given function:
So,
[tex]\begin{gathered} f(x)=3-x \\ f^{\prime}(x)=-1 \end{gathered}[/tex]Now,
Since f'(x) exists for all x, the only critical numbers of occure when f'(x) = 0.
Then,
According to the given differentiation, there are no critical points.
That means:
[tex]-1\ne0[/tex]Now,
The values at the endpoints of the interval, put -1 and 2 into the above function:
[tex]f(-1)=4[/tex]And,
[tex]\begin{gathered} f(x)=3-x \\ f(2)=3-2 \\ f(2)=1 \end{gathered}[/tex]Comparing these two numbers, we see the absolute maximum value and the absolute minimum values.
Final answer:
Hence, the minimum and maximum extrema are shown below:
[tex]\begin{gathered} minima:(2,1) \\ maxima:(-1,4) \end{gathered}[/tex]