First-Derivative Test: Finding Absolute Extrema (x=-8.8 and x=-7.2)

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In summary, the conversation discusses the concept of relative and absolute extrema, particularly in regards to the graph of a function. The critical points of the function, where g'(x) changes signs, are at x = -8.8 and x = -7.2. For x < -8.8 and x > -7.2, g'(x) is positive, while for -8.8 < x < -7.2, g'(x) is negative. It is determined that the point at x = -8.8 is a relative maximum, while the point at x = -7.2 is a relative minimum. The difference between local and absolute extrema is also discussed.
  • #1
Slimsta
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Homework Statement


http://img5.imageshack.us/img5/2218/49718059.jpg


Homework Equations


in picture


The Attempt at a Solution


why am i getting this wrong?
when x=-8.8 its the abs.max value
and x=-7.2 is the abs.min value...
 
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  • #2
-8.8 appears to be a relative min if g'(x) changes from - to +.

Think about it down and then up: \ /

seems like a minimum
 
  • #3
MathDude said:
-8.8 appears to be a relative min if g'(x) changes from - to +.

Think about it down and then up: \ /

seems like a minimum

im a bit confused with the wording.
relative = local or absolute?
i assumed it is absolute.

so i kinda get why -8.8 is min.. but then -7.2 has to be max, which would look like: /
is that correct?
 
  • #4
I had a great right up for you, but when I posted it I was timed out and had to login in (drives me nuts!) and lost everything. I don't have time to write it again, but here is a short version.

g'(x) is

+ (-inf, -8.8)
- (-8.8, -8)
- (-8, -7.2) Asymptote at -8
+ (-7.2, + inf)

I didn't observe the vertical asymptote at -8 in my earlier post. This changes the slope.

-8.8 is actually a max

-7.2 is a min.

Relative implies either local or abs. Evaluating g(x) at the critical points will enable you to tell if its local or abs. Abs values will yields the very largest and very smallest values for g(x) for all values of x within the given domain. Locals give you min or max within a local range. Locals can be absolute if there is only one max and/or one min. Say you have two rel max, the point with a larger g(x) will be the abs and the smaller g(x) will be the local.
 
  • #5
MathDude said:
I had a great right up for you, but when I posted it I was timed out and had to login in (drives me nuts!) and lost everything. I don't have time to write it again, but here is a short version.

g'(x) is

+ (-inf, -8.8)
- (-8.8, -8)
- (-8, -7.2) Asymptote at -8
+ (-7.2, + inf)

I didn't observe the vertical asymptote at -8 in my earlier post. This changes the slope.

-8.8 is actually a max

-7.2 is a min.

Relative implies either local or abs. Evaluating g(x) at the critical points will enable you to tell if its local or abs. Abs values will yields the very largest and very smallest values for g(x) for all values of x within the given domain. Locals give you min or max within a local range. Locals can be absolute if there is only one max and/or one min. Say you have two rel max, the point with a larger g(x) will be the abs and the smaller g(x) will be the local.


so would that work:
http://img29.imageshack.us/img29/5402/40917982.jpg

the only thing i don't get is why
- (-8.8, -8)
- (-8, -7.2)
and not +?
 
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  • #6
g'(x) can change signs only at zeros in the numerator, which are x = -8.8 and x = -7.2. For x < -8.8, g'(x) > 0. For x > -7.2, g'(x) > 0. For -8.8 < x < -7.2, g'(x) < 0 (although g'(x) is undefined at x = -8.)
 
  • #7
Mark44 said:
g'(x) can change signs only at zeros in the numerator, which are x = -8.8 and x = -7.2. For x < -8.8, g'(x) > 0. For x > -7.2, g'(x) > 0. For -8.8 < x < -7.2, g'(x) < 0 (although g'(x) is undefined at x = -8.)

i got it! the problem was with the + and - signs.. i had to switch places! what a stupid mistake :S
 
Last edited:

1. What is the First-Derivative Test?

The First-Derivative Test is a mathematical tool used to determine whether a critical point on a function is a relative maximum or minimum. It involves taking the first derivative of the function and evaluating it at the critical point.

2. How do you find the critical points using the First-Derivative Test?

To find the critical points using the First-Derivative Test, you first take the derivative of the function and set it equal to zero. Solve for the variable to find the x-values of the critical points. Then, plug these values into the first derivative to determine if they are relative maximums or minimums.

3. Can the First-Derivative Test be used for all functions?

No, the First-Derivative Test can only be used for differentiable functions. This means that the function must have a defined derivative at every point on its domain.

4. What is the difference between a relative maximum and a relative minimum?

A relative maximum is the highest point on a function within a specific interval, while a relative minimum is the lowest point within that interval. These points can also be referred to as local maximums and minimums.

5. Are there any limitations to using the First-Derivative Test?

One limitation of the First-Derivative Test is that it can only determine relative maximums and minimums, not absolute maximums and minimums. Additionally, it may not work for functions with discontinuities or sharp corners.

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