C3 Trig question

(a) Sketch the graph of y=1+2secx in the interval −π ≤ θ ≤ 2π.

(b) Write down the y-coordinate of points at which the gradient is zero.

(c) Deduce the maximum and minimum values of 1/ 1+2secx and give the smallest positive values of θ at which they occur.

If you want to know what part a) looks like without sketching for yourself, the sketch and answers are here.

https://activeteach-prod.resource.pearson-intl.com/r00/r0069/r006953/r00695312/current/alevelsb_p2_ex6b.pdf
I don't understand part c) at all tbh, could someone help me?

(edited 5 years ago)

Reply 1

begbie68

16

I'm guessing that you (or someone) has mixed 'x' with 'θ' in the question.

So, you've sketched the graph in part (a), and found the max/mins in part (b) .... which will occur at

first min : θ = 0, then 2pi
First max : θ = pi

So, for part (c), what's happening?
- they've transformed the graph by finding the reciprocal.

So, what was a BIG value in the original function (or graph) is now a SMALL value in the reciprocal graph, and vice versa.
eg
1/large = small
&
1/(very small) = very large.

Then you can see what the question is asking you now?

(edited 5 years ago)

Reply 2

dont know it

OP

9

Original post by begbie68

I'm guessing that you (or someone) has mixed 'x' with 'θ' in the question.

So, you've sketched the graph in part (a), and found the max/mins in part (b) .... which will occur at

first min : θ = 0, then 2pi
First max : θ = pi

So, for part (c), what's happening?
- they've transformed the graph by finding the reciprocal.

So, what was a BIG value in the original function (or graph) is now a SMALL value in the reciprocal graph, and vice versa.
eg
1/large = small
&
1/(very small) = very large.

Then you can see what the question is asking you now?

Not really tbh. I just don't get how the minimum/maximum points from 1+2secx translate to telling us the minimum and maximum for 1/1+2secx.

Original post by dont know it

Not really tbh. I just don't get how the minimum/maximum points from 1+2secx translate to telling us the minimum and maximum for 1/1+2secx.

But he mentioned this.

Essentially, the maximum of 1+2sec(x) will give you the minimum of 1/(1+2sec x) if you take the reciprocal of it.

Likewise, the minimum of 1+2sec(x) will give you the maximum of 1/(1+2sec x) if you take the reciprocal of it.

Genrally, this rule shouldn’t be applied blindly without more analysis, but it’s important that you understand the concept of when you have (large number) then the reciprocal 1/(large number) will give you something very small.
The largest possible number you can have will yield the smallest possible number you can have as I explained above.