MHB What Are the Absolute Max/Min of f(x) Over Given Intervals?

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To find the absolute maximum and minimum of the function f(x)=x^3-12x^2-27x+8 over specified intervals, the critical points at x=-1 and x=9 must be evaluated along with the endpoints of each interval. In the interval [-2,0], the maximum occurs at f(-1) and the minimum is not defined. For the interval [1,10], both the maximum and minimum are not defined as the function does not reach a peak or trough. In the interval [-2,10], the maximum is still at f(-1) and the minimum remains undefined due to the behavior of the function. It is essential to evaluate the function at both critical points and endpoints to determine the correct absolute extrema within the given intervals.
riri
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Hello, I'm doing:
Find absolute min/max of f(x)=x^3-12x^2-27x+8

First I found derivative and when I solved, I got x=9 and x=-1.
So I have to find max/min for 3 different intervals:

a) [-2,0]
And I thought absolute max=-1 and absolute min = none?

b) [1,10]
max= none
min= none

c) [-2,10]
max=-1?
min=none again because for x=9, it goes decrease and decrease so it's not min or max.

But NONE of these are right, can anyone help me what I am doing wrong and what they are asking for then?
thank you!
 
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$$f'(x)=3x^2-24x-27=3(x+1)(x-9)=0$$
so there is local extremum at $x=-1$ and $x=9$. Plug these values into $f(x)$ to find the values of the extremum. As $f(x)$ tends to $\pm\infty$ as $x$ increases/decreases without bound, there are no defined absolute minimum and absolute maximum.
 
riri said:
Hello, I'm doing:
Find absolute min/max of f(x)=x^3-12x^2-27x+8

First I found derivative and when I solved, I got x=9 and x=-1.
So I have to find max/min for 3 different intervals:

a) [-2,0]
And I thought absolute max=-1 and absolute min = none?

b) [1,10]
max= none
min= none

c) [-2,10]
max=-1?
min=none again because for x=9, it goes decrease and decrease so it's not min or max.

But NONE of these are right, can anyone help me what I am doing wrong and what they are asking for then?
thank you!

The absolute maximum or absolute minimum of a function over an interval can occur either at a turning point or an end point.

Also the maximum/minimum VALUE of the function is the y value, positioned where x = blah...

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greg1313 said:
$$f'(x)=3x^2-24x-27=3(x+1)(x-9)=0$$
so there is local extremum at $x=-1$ and $x=9$. Plug these values into $f(x)$ to find the values of the extremum. As $f(x)$ tends to $\pm\infty$ as $x$ increases/decreases without bound, there are no defined absolute minimum and absolute maximum.

Except that the OP needs to find the absolute max/min of the function IN THE INTERVALS PROVIDED.
 
riri said:
Hello, I'm doing:
Find absolute min/max of f(x)=x^3-12x^2-27x+8

First I found derivative and when I solved, I got x=9 and x=-1.
So I have to find max/min for 3 different intervals:

a) [-2,0]
And I thought absolute max=-1 and absolute min = none?

b) [1,10]
max= none
min= none

c) [-2,10]
max=-1?
min=none again because for x=9, it goes decrease and decrease so it's not min or max.

But NONE of these are right, can anyone help me what I am doing wrong and what they are asking for then?
thank you!
check this out and let me know if you have further questions, I am here to help