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I'm having trouble with a question. The question is:

At what value of x does the graph of y= $x^{-4}-x^{-5}$ have a point of inflection? Discuss the concavity of the graph.

I got a POI at $\frac{3}{2}$, but apparently that is wrong, but no corrections were given so I am confused on how to solve this problem the correct way and get another answer.

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I get $y''=10x^{-7}(2x-3)$ which confirms your answer! – Maesumi Jan 24 '13 at 4:36
Captn Buzz: you may accept one answer per question , if helpful, by clicking on the "greyed out" "checkmark" to the left of the question. Once you get 30 rep, you can also upvote as many answers to questions as you like. – amWhy Jan 24 '13 at 20:19
up vote 2 down vote accepted

Our function is $x^{-4}-x^{-5}$. Differentiate. We get $-4x^{-5}+5x^{-6}$. Again. We get $20x^{-6}-30x^{-7}$. This indeed is $0$ at $x=\frac{3}{2}$, and nowhere else.

The second derivative can be written as $\frac{20x-30}{x^7}$.

When $x$ is negative, the second derivative is positive, our function is "concave up" (this is the usual terminology in North American calculus books).

When $x$ is positive but less than $\frac{3}{2}$, the second derivative is negative, the curve is concave down.

Finally, for $x\gt \frac{3}{2}$ the curve is concave up.

Remark: Without seeing what you wrote, it is not possible to know why you might have been downgraded. You identified correctly the point of inflection. It is possible that you missed the change in concavity at $x=0$.

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