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January 30, 2015

January 30, 2015

Posted by **Anonymous** on Tuesday, June 1, 2010 at 6:30pm.

a. increasing

b. decreasing

c. concave up

d. concave down

Then find any

e. local extreme values

f. inflection points

---

I thought it'd be better to find extreme values first, then you can find the points where the function is increasing or decreasing. (I'll worry about concavity and inflection points later, since that's second derivative)

So I found y':

x^3-2x^2-x

...and set it equal to zero to find the points where derivative crosses the x-axis

x^3-2x^2-x=0

I factored and used the quadratic formula, and got (2+sqrt8)/2 and (2-sqrt8)/2, or -.414 and 2.414. Oh, and zero too because I factored out an x.

So my problem is that the answers (which were given to me ahead of time to check my work) don't match my work. Apparently y is increasing from [0, infinity) and decreasing from (-infinity, 0], and none of the points I found from finding y'.

Can you find what I did wrong?

Thanks!

- Calculus -
**Reiny**, Tuesday, June 1, 2010 at 8:09pmYour derivative is wrong,

you should have

y' = x^3 - 2x^2 + x

When I set x^3 - 2x^2 - x = 0, I get

x(x^2 - 2x + 1 ) = 0

x(x-1)(x-1) = 0

x = 0 or x = 1

take it from there.

- Calculus -
**Anonymous**, Tuesday, June 1, 2010 at 8:12pmOh wow, such a simple mistake throws off everything, haha. Thanks

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