How am I meant to deduce the range of values for concavity here ?

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seals2001
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I have calculated x = 0.322 and 1.89 when f ' ' (x) = 0.

But from the graph, what do I do to determine the range of values which f(x) is concave ? You can't just look at the graph to do this, it has to be done algebraically, so how do I determine the range of values which tan(2x) > 0 ?

Thanks.
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DFranklin
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You should be able to find f''(x) in the form (A cos 2x + B sin 2x) / e^x. You want the values where f''(x) > 0, which you can do by converting A cos 2x + B sin 2x into the R \sin(\theta + \alpha) form.

Edit: finding where tan 2x > 0 won't give you the correct answer. At the same time, it's somewhat worrying if you can't solve this.
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seals2001
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(Original post by DFranklin)
You should be able to find f''(x) in the form (A cos 2x + B sin 2x) / e^x. You want the values where f''(x) > 0, which you can do by converting A cos 2x + B sin 2x into the R \sin(\theta + \alpha) form.

Edit: finding where tan 2x > 0 won't give you the correct answer. At the same time, it's somewhat worrying if you can't solve this.
Why wont tan(2x) give me the correct answer when I want to find f ' ' (x) > ) ? Also obviously I can solve it graphically, but that is not practical for a 2 mark question. How can I solve it quickly algebraically ?
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ghostwalker
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(Original post by seals2001)
Why wont tan(2x) give me the correct answer when I want to find f ' ' (x) > ) ? Also obviously I can solve it graphically, but that is not practical for a 2 mark question. How can I solve it quickly algebraically ?
You were looking for tan(2x)>0, which is wrong on two accounts.

Your 2nd deriviative required 4\sin 2x - 3\cos 2x\geq 0

For your f''(x)=0 points I presume you solved 4\sin 2x - 3\cos 2x= 0 by rearranging it to \tan 2x = 3/4. That's fair enough. So, you might think to look for tan(2x)>3/4 rather than 0.

However, in going from 4\sin 2x - 3\cos 2x\geq 0 to \tan 2x \geq 3/4 you wil have divided by \cos 2x. But \cos 2x isn't always positive, and will flip sign at intervals of \pi/2. And you know if you divide by a negative value the inequality reverses. So, you'd end up having to split the x-axis up into intervals of \pi/2 with the inequality being the other way around in consecutive intervals. Nightmare!

You'll also have introduced an issue with tan(2x) not being defined when x=pi/4 (amongst other places) which is in the interval you're interested in.


Part (c) is worth 5 marks. You'd get most of them for finding the points where f''(x) is 0. Then refering to the graph, you can see that it will be concave between those two points, and convex outside that interval.

In order to change from concave to convex, because f''(x) is continuous, you have to go through a point where f''(x) is zero, and that's why a quick confirmation checking the graph is all that's needed, IMO.

If you want an algebraic confirmation, then what DFranklin said is the way to go.
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