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Question 1001656: Let f(x) = x^4 + 2x^3 + 1
I take the derivative and set equal to zero to find its critical pts: x= 0 and x = -3/2. Then found the graph from decreasing(-∞,-3/2]
and increasing [-3/2,0]U[0,∞) "by the way are these brackets correct? is that correct to include the critical pts in the interval?"
Found that the rel extreme is therefore a MIN when x=-3/2
Found its concavity intervals and inflections pts at: x=-1 and x=0.
Concave up: (-∞,1)U(0,∞)
Concave down: (-1,0)
Found the yint at (0,1) but couldn't find the yint. And I guess these cannot actually have H.A. or V.A. that's only for rational functions.
I was able to sort of graph it
here is my graph: http://imgur.com/mFKoUy3
just confused, does the inflection point represent not only where the sign change happens but also a point on which the graph crosses the x-axis? And how does one find the x points from that function to get a complete scope of the graph. I cannot use a calculator.
Thank you!
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x^4 + 2x^3 + 1
derivative =0 is 4x^3+6x^2=0. Your critical points are correct. I would use open brackets, because at the critical point for a minimum, the graph is instantaneously not moving in the vertical direction. I would keep it open for the inflection points, too.
f''(x)=12x^2+12x, and I agree with the inflection points.
y-intercept without a calculator.
x^4+2x^3+1=0
(x+1)(x^3+x^2-x+1)=0
By synthetic division, x= -1 is a root, so (x+1) is a factor. The other function does not have an integer root. Minus 2 is almost a root but isn't.
The function goes to +oo for large minus x and large +x.
The y-intercept is (0,1)
There are two real roots, -1 and the other needs a calculator, but it is between -1.5 and -2 (it is -1.83)
The inflection point does occur where the graph crosses the x-axis.
You have a function that goes to positive infinity rapidly for positive and negative x.
The y-intercept and roots are known.
The only change I would make is that the inflection point is where the graph starts rising rapidly, rather than rising more slowly. You know that, because the first derivative is large negative or large positive, so you have almost a vertical line when x is not very large.
The minimum is known, (-3/2,-11/16)
While you can't use a calculator, for learning purposes, it's worth seeing the graph anyway.
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