On what interval/s is the function 9xtan(x) concave up? Please help! Thank you!
- Thread starter Lolaploca
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Hello, and welcome to FMH!
Can you state what we need in order to analyze a function's concavity?
Hello, thank you for providing us with this great help!
In order to analyze that we need to know the function's second derivative, and inflection points.
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Yes, good.
Have you computed the given function's 2nd derivative? If so what did you get?Yes, good.
I got:
f''(x) = 9[2sec^2(x)+2xsec^2(x)tan(x)]
I can't go farther than this.
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Okay, good...I get:
[MATH]f''(x)=18\sec^2(x)(x\tan(x)+1)[/MATH]
which is equivalent to your result. And so we want to find the roots of this 2nd derivative. We can ignore the 18, and so I would first think about:
[MATH]\sec(x)=0[/MATH]
Are there any real solutions to this equation?
[MATH]f''(x)=18\sec^2(x)(x\tan(x)+1)[/MATH]
which is equivalent to your result.
And so we want to find the roots of this 2nd derivative. We can ignore the 18, and so I would first think about:[MATH]\sec(x)=0[/MATH]
Are there any real solutions to this equation?
Okay, good...I get:
[MATH]f''(x)=18\sec^2(x)(x\tan(x)+1)[/MATH]
which is equivalent to your result.
[MATH]\sec(x)=0[/MATH]
Are there any real solutions to this equation?
Is there no solution?
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Right, and because it is squared, that factor will be positive for all \(x\), and so we really only need to be concerned with the sign of the remaining factor:
[MATH]g(x)=x\tan(x)+1[/MATH]
Before we look into finding the roots, I would observe that this function is even and has vertical asymptotes at:
[MATH]x=\frac{\pi}{2}(2k+1)[/MATH] where \(k\in\mathbb{Z}\)
Further, we find:
[MATH]g'(x)=\frac{\sec^2(x)}{2}(2x+\sin(2x))[/MATH]
And this derivative is positive for all positive \(x\), which means \(g(x)\) must have a single root in between each adjacent pair of asymptotes. Given that:
[MATH]g(0)=1[/MATH]
then we know:
[MATH]g(x)>0[/MATH] on [MATH]\left(-\frac{\pi}{2},\frac{\pi}{2}\right)[/MATH]
Now, let's call the root \(r_n\) on the interval [MATH]\left(\frac{\pi}{2}(2n+1),\frac{\pi}{2}(2n+3)\right)[/MATH] where \(n\in\mathbb{N}\).
We then know that \(g(x)>0\) on:
[MATH]\left(r_n,,\frac{\pi}{2}(2n+3)\right)[/MATH]
And by the even symmetry we also know that \(g(x)>0\) on:
[MATH]\left(-\frac{\pi}{2}(2n+3),-r_n,\right)[/MATH]
To find a particular \(r_n\), it will be necessary to use a numeric root finding technique, such as the Newton-Raphson method. Here are a few of the roots:
which I found here:
x*tan(x)+1=0 - Wolfram|Alpha
Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.
www.wolframalpha.com
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Wow, that's so helpful, thank you! I think I understand the problem better now.
Am I missing something? I can't see what the Concave Up intervals really are.
Here is a diagram showing some of the intervals where concavity is up, and are are shaded in blue:
Each of the dotted lines represents a root of \(g(x)\) which can only be approximated via a numeric root finding technique.