Factorising has nothing to do with locating maxima/minima. You would complete the square to find the maximum/minimum of a quadratic (this is basic C1 knowledge, but not sure what stage you're at in school).

And yes it's a positive quadratic but I did mention that

y

i.e.

\sin x

is restricted to a certain interval that you know, so it has a maximum.

Reply 8

Carman3

OP

12

Original post by IrrationalRoot

Factorising has nothing to do with locating maxima/minima. You would complete the square to find the maximum/minimum of a quadratic (this is basic C1 knowledge, but not sure what stage you're at in school).

And yes it's a positive quadratic but I did mention that

y

i.e.

\sin x

is restricted to a certain interval that you know, so it has a maximum.

Can you show me how to do it i cant get it

Reply 9

IrrationalRoot

20

Original post by Carman3

Can you show me how to do it i cant get it

You're trying to maximise the function

2y^2-y+\frac{1}{3}

with the restriction that

-1 \leq y \leq 1

.

So first, complete the square. Then think about which value of

y

between

-1

and

1

(inclusive) is going to maximise the function.

Reply 10

Carman3

OP

12

Original post by IrrationalRoot

You're trying to maximise the function

2y^2-y+\frac{1}{3}

with the restriction that

-1 \leq y \leq 1

.

So first, complete the square. Then think about which value of

y

between

-1

and

1

(inclusive) is going to maximise the function.

ok i get it thanks. how do you know if they are talking about the max/min value which would be the vertex or the maximum/min overall value like the above

Reply 11

IrrationalRoot

20

Original post by Carman3

ok i get it thanks. how do you know if they are talking about the max/min value which would be the vertex or the maximum/min overall value like the above

It's not about which one they're talking about, it's about what the function is.

2\sin^2 x - \sin x + \frac{1}{3}

is not the same function as

2x^2-x+\frac{1}{3}

(which is a parabola with a vertex and such). It's just that the maximum of

2\sin^2 x - \sin x + \frac{1}{3}

is going to be the same as that of the function

2x^2-x+\frac{1}{3}

restricted to

-1 \leq x \leq 1

, since sine can only take those values.

Reply 12

some-student

7

Original post by Carman3

What is the maximum value of 2sin^x - sinx + 1/3

To get the maximum value:
we want the maximum value of

2\mathrm{sin}^2 \, x

we want the minimum value of

\mathrm{sin} \, x

as we are subtracting it from our maximum of

2\mathrm{sin}^2 \, x

we cannot change

\frac{1}{3}

as it is a constant

Solution

Spoiler

Reply 13

the bear

20

i would use differentiation :dontknow:

Reply 14

IrrationalRoot

20

Original post by some-student

To get the maximum value:
we want the maximum value of

2\mathrm{sin}^2 \, x

we want the minimum value of

\mathrm{sin} \, x

as we are subtracting it from our maximum of

2\mathrm{sin}^2 \, x

we cannot change

\frac{1}{3}

as it is a constant

Solution

Spoiler

Yep good approach for this particular problem but OP should remember that this method does not work in general for similar functions.

Reply 15

Carman3

OP

12

Original post by some-student

To get the maximum value:
we want the maximum value of

2\mathrm{sin}^2 \, x

we want the minimum value of

\mathrm{sin} \, x

as we are subtracting it from our maximum of

2\mathrm{sin}^2 \, x

we cannot change

\frac{1}{3}

as it is a constant

Solution

Spoiler

Thanks...
Dont answer the question from the paper as i havent tried it yet but would you also do the same for this: and similar questions as well

https://s3-eu-west-1.amazonaws.com/www.tutorfair.com/resources/attachments/d3a7c8252adc8fd22abb6bc1a79d877756dfc5cd_1.pdf

Question 1E

Reply 16

some-student

7

Original post by IrrationalRoot

Yep good approach for this particular problem but OP should remember that this method does not work in general for similar functions.

Yeah - I should comment that this problem only happens as when

\mathrm{sin}x = -1

, it lines up with

\mathrm{sin}^2x = 1

, and this isn't always the case.

Reply 17

RogerOxon

21

Original post by IrrationalRoot

You're trying to maximise the function

2y^2-y+\frac{1}{3}

with the restriction that

-1 \leq y \leq 1

.

So first, complete the square. Then think about which value of

y

between

-1

and

1

(inclusive) is going to maximise the function.

I don't see a need to complete the square here, although it is a technique that you need to know and is useful for sketching the function.

As it's a parabola with a positive

y^2

coefficient, you know that the maximum value over any range is going to be at the end of the range. For a symmetric range, the negative

y

coefficient tells you which end, or simply trying the two values.

(edited 7 years ago)

Reply 18

IrrationalRoot

20

Original post by RogerOxon

I don't see a need to complete the square here, although it is a technique that you need to know and is useful for sketching the function.

As it's a parabola with a positive