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Estimate the intervals of concavity to one decimal place by using a computer algebra system to compute and graph $ f" $.

$ f(x) = \frac{x^4 + x^3 + 1}{\sqrt{x^2 + x + 1}} $

From the graph above, we see that $f(x)$ is

concave upward on $(-\infty,-0.62) \cup(0.02, \infty)$ and

concave downward on $(-0.62,0.02)$

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Missouri State University

Campbell University

Harvey Mudd College

Idaho State University

So I've used a computer system to find the second derivative of our original function F of X. And it came out to be this huge, very complicated function. Um Which would have been very hard to find if we didn't use a computer. Um And so now we're just gonna look at the graph which I have here on dez mose. And we're gonna look at what the intervals of con cavity are. So where this is concave up and concave down. Well, we know we're concave up whenever we have positive Y values for our second derivative. So that's going to be from negative infinity. Since it doesn't look to be coming back down anytime soon. To this point here at- .635. And then We go where concave down from negative .635 all the way to this point at .029. And then we're back to being concave up. So The points to remember are negative .635 and negative .029. So if you go back here can say that we are concave up from negative infinity- .635. And from 0.29 to infinity. Since those where we had positive values for our second derivative and we are concave down on the interval of negative point. Did I not put a negative sign up here Should be -635. Um Yeah, so from negative .635 2.029 is where we are con caved out. So these are the two intervals for concave up and concave down. And we found these just using the graph of our second derivative.