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January 30, 2015

January 30, 2015

Posted by **unknown** on Sunday, October 2, 2011 at 5:02am.

find the:

a) critical value(s)

b) critical point(s)

c) max. value + max. point

d) min.value and min. point

e)point on inflection if there is:

1) y=x^4-3x^3+22x^2-24x+12

thanks:)

- Diff Calculus -
**Reiny**, Sunday, October 2, 2011 at 8:44amA rather uninspiring graph

There is an obvious y-intercept of 12

after trying the usual methods of finding some f(a) = 0

I used WolframAlpha to see what we had

http://www.wolframalpha.com/input/?i=x%5E4-3x%5E3%2B22x%5E2-24x%2B12+

It showed no x-intercepts, thus no solution for f(x) = 0 , but rather 4 complex roots

I then clicked on the derivative expression to get one real root at appr x = .6

f(.6) = .6^4 - 3(.6)^3 + 22(.6)^2 - 24(.6) + 12 = appr. 5

So there is a turning point at about (0.6 , 5)

see http://www.wolframalpha.com/input/?i=-24%2B44+x-9+x%5E2%2B4+x%5E3&lk=1

clicking on the derivative of that cubic shows that the resulting quadratic has no real roots, thus no points of inflection

(It would have been a nightmare to attempt this without some convenient software, solving this with only pencil and paper and a basic calculator seems daunting )

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