 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: 2,553 Thanks: 1403 $f(x) = -2x^6 + 15x^5$ $f^{\prime\prime}(x) = -60x^4 + 300x^3$ $f^{\prime\prime}(x) = 0 \Rightarrow 60x^4 = 300x^3$ 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: 20,972 Thanks: 2222 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: 20,972 Thanks: 2222 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 $x< 0 \Rightarrow f^{\prime\prime}(x)<0 \text{ and }x > 0 \Rightarrow f^{\prime \prime}(x)>0$ Tags derivatives, finding inflection points, math, mikeiketube, problems Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MikeAndIke Calculus 1 February 18th, 2019 03:44 PM MikeAndIke Calculus 2 February 18th, 2019 07:44 AM johnmath Math Books 6 January 27th, 2013 03:13 PM SedaKhold Calculus 0 February 13th, 2012 11:45 AM katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM

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