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February 18th, 2019, 03:28 PM   #1
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Exclamation "Finding Inflection Points" Math Problems Help 4

Let f(x)=-2x^6+15x^5.
For what values of x does the graph of f have a point of inflection?

I found the second derivative and when factored leaves me with: -60x^3(x-5).
So, I figured the answer is just x=5 but my friend told me its zero as well, but how?

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February 18th, 2019, 03:55 PM   #2
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$f(x) = -2x^6 + 15x^5$

$f^{\prime\prime}(x) = -60x^4 + 300x^3$

$f^{\prime\prime}(x) = 0 \Rightarrow 60x^4 = 300x^3$

This has an obvious solution of $x=0$

$x \neq 0 \Rightarrow x = 5$

So the two inflection points are at $x = 0$ and $x=5$
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February 18th, 2019, 04:21 PM   #3
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But how exactly is it zero?
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February 18th, 2019, 04:31 PM   #4
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That reasoning, romsek, wouldn't suffice for, say, $f(x) = -2x^7 + 14x^6$.
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February 18th, 2019, 05:01 PM   #5
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Having found that $f\,''(x) = 0$ for some value of $x$, one can evaluate successive higher derivatives at the same value of $x$ until one finds a higher derivative that is non-zero there. If that's an odd-order derivative, there is a point of inflection there.

If the function has all its higher derivatives equal to zero, the issue still isn't decided.

There are also functions such as $f(x) = x^{1/3}$ that have a tangent that crosses the curve at its point of contact with the curve, but aren't differentiable there.
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February 18th, 2019, 05:05 PM   #6
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Quote:
Originally Posted by skipjack View Post
That reasoning, romsek, wouldn't suffice for, say, $f(x) = -2x^7 + 14x^6$.
You're correct.

I omitted the part where you must check for a sign change in any epsilon region about the stationary point.

It's not hard to see that $x< 0 \Rightarrow f^{\prime\prime}(x)<0 \text{ and }x > 0 \Rightarrow f^{\prime \prime}(x)>0$
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