Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use.

f(x)=9x^3−27x+9, [0,3]

Question

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use.

f(x)=9x^3−27x+9, [0,3]

anonymous · Answer

i know that you have to find the directive 
that is 
27^2-27

anonymous · Answer

I keep getting the wrong answer 
(1,0) and (-1,0)
the right answer is 
(3,171) and (1,-9)
I just don't know how to get that to work out

zehanz · Answer

The derivative (not directive ;)) is$$f'(x)=27x^2-27$$$$27x^2-27=0 \Leftrightarrow x^2-1=0 \Leftrightarrow x^2=1 \Leftrightarrow x= \pm 1$$
You have also found these x-values, but you plugged them in f', which will give you 0.
You should put these x-values in the original function.

anonymous · Answer

that gave me -9 and 27

zehanz · Answer

$$f(-1)=9(-1)^3-27 \cdot -1 + 9=-9+27+9=27$$$$f(1)=9 \cdot 1^3 -27 \cdot 1+9=9-27+9=-9$$Rembember, these outcomes, 27 and -9 are -at this moment- only the y-coordinates of points of the graph of f where there is a horizontal tangent line (f'= 0). 
We do not know if they are minimum of maximum values.
Because the domain is [0, 3], the graph begins in x=0 and ends at x=3. 
You also have to know what f(0) and f(3) are.

anonymous · Answer

I got 9 and 171

zehanz · Answer

f(0)=9
$$f(3)=9 \cdot 3^3 -27 \cdot 3 + 9=171$$
This is the graph:
So: x = -1 is irrelevant: ouside of [0,3].
Absolute minimum for x = 1, value -9
Absolute maximum for x = 3, value 171

anonymous · Answer

where did you get the number 1

zehanz · Answer

Although f(0)=9, 9 has no special meaning. it is just an in-between value.
They ask you for the absolute minumum and the abs max.
Because graphs go up and down, the points with a horizontal tangent line are interesting: with this function, you got a horizontal tangent line for x = 1. If you look at the graph, you'll see that for x=1, the graph is at its lowest point (-9). The highest point is not where the tangent line is horizontal, it's just where the quickly ascending graph is cut off because it goes no further than x=3. Value there: 171.

zehanz · Answer

You'll have to realize that this graph is only part of the complete graph: it's the part that goesfrom x=0 to x=3. See image. You can also see the two values, -1 and , where the tangent line is horizontal. For x=-1 there is a (local) maximum, for x=1 there is a (local) minimum.