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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(x5). 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 $x=5$ 
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, $f(x) = 2x^7 + 14x^6$.

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 nonzero there. If that's an oddorder 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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  Quote:
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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