A bit perplexed..

[legacyofpiracy]

Alright so I am having quite a bit of trouble with this particular section, possibly because our teacher whipped right through it, and the explinations in the book are simply making me more confused. Generally I am having difficulties finding the extreme values of a function without the use of a calculator. There are quite a few mismatched steps floating about in my mind without any set order, such as when to find derivative of the function, set it equal to zero, something about the denominator having to equal zero, etc....as you can see I am a bit lost so I am looking for any bit of help anyone has to offer.

For example, a homework problem asks me to find the extreme values of the function and where they occur:

Y= x^3+x^2-8x+5

I am not sure where to begin but I do know that the derivative is 3x^2+2x-8 if that helps at all :\


-EDIT-

I am also similarily confused on the use of the mean value theorem in an equation such as

f(x)=5x-x^2

where I am asked to find the local extreems and determine the intervals where the func is increasing and decreasing


If someone could possibly walk me through the process of solving you have no idea how much I would appreciate it.

 

[Guest]

ok to start with the part you can do.....3x^2+2x-8
You have found the y' = 3x^2+2x-8
This is the equation of the slope or gradient at any point on your original graph. You let this equal zero when you want no slope or a horizontal line. This is the max. or min. point.

y' = 3x^2+2x-8
0 = 3x^2+2x-8
and solve this --it will give you the "x" point where the slope = 0. Then put the x point back into the original equation to find the matching "y" point.

Comments?

 

[legacyofpiracy]

Wow thank you for replying so fast

Just one question though..

When you solve for 0 = 3x^2+2x-8 you factor correct? Then wouldn't this give you two values...so which value would you plug back into the original equation to get the matching y value?

if that was unclear i appologize, let me know if I should re-word

 

[Daniel_Feldman]

You have to do what's called a "factor line analysis." After getting the two values, plot them on a number line and find the intervals where the derivative is positive and the intervals where it is negative. The value at which the derivative changes from + to - is where there is a relative maximum, and the value at which the derivative changes from - to + is where there is a relative minimum. Plug in these values into the original function to get the respective max/min values.

 

[legacyofpiracy]

So would this be correct:

I got the two values -2, and 4/3 when I factored out the derivative 3x^2+2x-8 and I placed them on the numberline. By plugging in values greater, less, and between these two values i determined the interval less than -2 to be positive, between -2 and 4/3 to be negative, and greater than 4/3 to be positive.

 

[Daniel_Feldman]

Sorry for the delayed reply. Your work is absolutely correct. Now, what conclusions can you draw from it?