Critical points, minimum, maximum, and inflection points (NEW QUESTION)

[EETman]

Let g(x)=x⁹*e^x
my first differentiation reveals ((9x^8)+x^9)e^x, if i factor it further i can pull x^8 out and have only 9+x, however from the graph it doesnt seem that there is a critical point there, just one inflection point at 0,0. Question from this part is as such; am i looking for critical points that donot exist in this problem?

the second derivation revealed (x⁹+18x⁸+27x²)e^x , the graph of which appears parabolic in nature with an inflection point again at 0,0.

The problem is worded as follows " Determine the intervals where g is increasing and decreasing, concave up and concave down. Also, find all local minima/maxima and any inflection points. If there is neither a local max or a local mind, write none"

This problem has me perplexed mainly because we did not go over problems this "complicated" in class so im having some troubles finding out where to go from here, if you need any additional information feel free to ask, thanks.

 

[Romsek]

Let g(x)=x⁹*e^x
my first differentiation reveals ((9x^8)+x^9)e^x, if i factor it further i can pull x^8 out and have only 9+x, however from the graph it doesnt seem that there is a critical point there, just one inflection point at 0,0. Question from this part is as such; am i looking for critical points that donot exist in this problem?

the second derivation revealed (x⁹+18x⁸+27x²)e^x , the graph of which appears parabolic in nature with an inflection point again at 0,0.

The problem is worded as follows " Determine the intervals where g is increasing and decreasing, concave up and concave down. Also, find all local minima/maxima and any inflection points. If there is neither a local max or a local mind, write none"

This problem has me perplexed mainly because we did not go over problems this "complicated" in class so im having some troubles finding out where to go from here, if you need any additional information feel free to ask, thanks.

It's there at -9. You're just not seeing it on the function due to the scale. g'(x) has 2 zeros, first order at -9, and 8th order at 0. The 2nd derivative at -9 is positive and thus this is a local minimum.
At 0 it's not so easy since the derivatives there just keep being zero down the line. Since the 2nd derivative does exist at 0 and is neither positive nor negative g is neither concave upward or downward at 0.
An inflection point is a point where the 2nd derivative changes sign. It is necessarily a point where the 2nd derivative = 0, but that's not sufficient, you need to check for the sign change as well.
At 0 the 2nd derivative does in fact change sign and so this would be an inflection point.

Basically just use the rules given in your text for testing for minima, concavity up and down, and inflection points. It's all a matter of finding the zeros of the first and second derivatives and checking signs.

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[EETman]

So what of the inflection points?

 

[Romsek]

So what of the inflection points?

reread my post.