A 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 )
Find the interval(s) where the function is increasing of decreasing.
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:)
1 answer