My Math Forum "Finding Inflection Points" Math Problems Help #2

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 February 18th, 2019, 03:28 PM #1 Newbie     Joined: Feb 2019 From: N/A Posts: 18 Thanks: 0 Math Focus: Calculus "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? Last edited by MikeAndIke; February 18th, 2019 at 04:16 PM.
 February 18th, 2019, 03:55 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276 $f(x) = -2x^6 + 15x^5$ $f^{\prime\prime}(x) = -60x^4 + 300x^3$ $f^{\prime\prime}(x) = 0 \Rightarrow 60x^4 = 300x^3$ This has an obvious solution of $x=0$ $x \neq 0 \Rightarrow x = 5$ So the two inflection points are at $x = 0$ and $x=5$ Thanks from topsquark
 February 18th, 2019, 04:21 PM #3 Newbie     Joined: Feb 2019 From: N/A Posts: 18 Thanks: 0 Math Focus: Calculus But how exactly is it zero?
 February 18th, 2019, 04:31 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 That reasoning, romsek, wouldn't suffice for, say, $f(x) = -2x^7 + 14x^6$. Thanks from MikeAndIke
 February 18th, 2019, 05:01 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 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.
February 18th, 2019, 05:05 PM   #6
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Quote:
 Originally Posted by skipjack 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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