Finding the smallest value of f(x) = tan x + cot x

[sqleung]

Sorry to bother you guys again. You all have been very helpful but unfortunately, I have another question that I need assistance with. This question comes in two parts. I've managed to do the first one (whether it's right, I'm not too certain...) but the second one definitely confused me. I'm not too certain on the wording of the question either:

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1. Express f(x) = tan x + cot x in terms of sin 2x

This is my working out for this one:

(sinx / cosx) + (cosx / sinx)
(sin²a / cosxsinx) + (cos²x / sinxcosx)
(sin²x + cos²x) / cosxsinx
1 / cosxsinx
2 / 2cosxsinx
2 / sin2x

2. Find the smallest value of f(x) in the domain of 0 < x < ?/2. Also find the corresponding exact value for x. Justify your answer algebraically by using your answer to the first part

This is the one I'm stuck on. The wording of it threw me off completely. Can somebody perhaps shed a little light on this and translate it into "a language I can understand ". Also, would you be able to provide a starting point because I'm really terrible at finding those darn things! Alternatively, you could show me how to do it and I'll try and interpret it so I'll be prepared for any similar questions in the future.

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Thankyou. Of course, any help is appreciated.

 

[galactus]

Re: Finding the smallest value of f(x)

Good work on the first part.

For the second part, look at the graph of \(\displaystyle \frac{2}{sin(2x)}=2csc(2x)\)

You could try taking the derivative of 2csc(2x) to find the minimum in that interval.

Remember, \(\displaystyle \frac{d}{dx}[csc(x)] = -csc(x)cot(x)\)

Therefore, \(\displaystyle \frac{d}{dx}[2csc(2x)}=-4csc(2x)cot(2x)\)

\(\displaystyle \frac{-4}{sin(2x)}\cdot\frac{cos(2x)}{sin(2x)}=0\)

\(\displaystyle cos(2x)=0\)

Since \(\displaystyle cos(\frac{\pi}{2})=0\), what should be the value of x?.

 

[pka]

Re: Finding the smallest value of f(x)

Look carefully at the graph.
Where does the low point occur?
What is the value for the function there?

 

[skeeter]

I think there was a reason for having you do the first part ... use the result to think about the second question.

f(x) = tan(x) + cot(x) = 2/sin(2x), f(x) will have its minimum value when the denominator, sin(2x), is at its greatest value ... since the greatest value for sin(anything) = 1,
sin(2x) = 1 when x = pi/4

 

[sqleung]

A big thanks to all of you!

You've all been VERY helpful.