Methods Question

[Lear]

Is the approach that you have demonstrated above the same when you find the domain over which the gradient is negative?, as in whether it's ) or ]
?

Yep, sub in a value and observe whether the gradient is negative, positive or 0.

[secretweapon]

Yep, sub in a value and observe whether the gradient is negative, positive or 0.

Thanks
But when i substituted pi/4 into the derivate of the equation 3sin(2x) - 4, i got 3√(2) = ~4.24. But how come even though 4.24 >0, it's ) for pi/4?
[0, pi/4)

[Lear]

Thanks
But when i substituted pi/4 into the derivate of the equation 3sin(2x) - 4, i got 3√(2) = ~4.24. But how come even though 4.24 >0, it's ) for pi/4?
[0, pi/4)

I dont know what you are subbing in but i am getting 0 as the gradient at pi/4.

[secretweapon]

I dont know what you are subbing in but i am getting 0 as the gradient at pi/4.

Whoops my bad, accidentally subbed in the wrong number 

[secretweapon]

can the period of a function ever be negative? (like if you had y = sin(-x)

[S200]

can the period of a function ever be negative? (like if you had y = sin(-x)

No. The - sign would be a reflection, rather than the actual dilation factor.
It is the dilation factor (in your example 1) that is used while calculating the period.

[secretweapon]

No. The - sign would be a reflection, rather than the actual dilation factor.
It is the dilation factor (in your example 1) that is used while calculating the period.

Thanks
Another question, does it matter which order you do the dilation and reflection in a transformation matrix?

[S200]

I don't think so, but I really suck at transformations through matrices, so I can't be 100% sure.

[secretweapon]

Is anyone certain as to whether the order you do the dilations and reflections in transformation matrices matter?
Also, how would you describe the transformations that take -(2*x-2)^(1/2)+2 to (x)^(1/2)
the first equation is -(square root of 2x-2) + 2
and the second equation is square root of x
Thanks

[secretweapon]

Is anyone certain as to whether the order you do the dilations and reflections in transformation matrices matter?
Also, how would you describe the transformations that take -(2*x-2)^(1/2)+2 to (x)^(1/2)
the first equation is -(square root of 2x-2) + 2
and the second equation is square root of x
Thanks

Bump
Also, does anyone know how to find the anti derivative of 2sin(x) + k
Thanks

[Springyboy]

Bump
Also, does anyone know how to find the anti derivative of 2sin(x) + k
Thanks

So the anti-derivative of sin(x) = -cos(x)
Therefore, it'd just be -2cos(x)+k^2/2

As the anti-derivative of k = k^2/2

[secretweapon]

So the anti-derivative of sin(x) = -cos(x)
Therefore, it'd just be -2cos(x)+k^2/2

As the anti-derivative of k = k^2/2

Thanks
I was finding the anti derivative of 0.25x^-2
here is my working
(0.25x^-1)/(0.25*-1)
= (0.25x^-1)/(-0.25)
but the answer said it was -1/4x
Do you know what is incorrect in my working out?

[Springyboy]

Thanks
I was finding the anti derivative of 0.25x^-2
here is my working
(0.25x^-1)/(0.25*-1)
= (0.25x^-1)/(-0.25)
but the answer said it was -1/4x
Do you know what is incorrect in my working out?

So with this I'd rewrite out 0.25x^-2 to 1/4x^2 as remember that 0.25 = 1/4

That way you're taking the anti-derivative of 1/4x^2 = 1*4x^-2 = -1*(1/4x^-1) = -1/4x which is the answer given

That's due to 0.25 = 1/4 so you just need to rewrite 0.25 as 1/4 to simplify the process and get the right answer

[secretweapon]

So with this I'd rewrite out 0.25x^-2 to 1/4x^2 as remember that 0.25 = 1/4

That way you're taking the anti-derivative of 1/4x^2 = 1*4x^-2 = -1*(1/4x^-1) = -1/4x which is the answer given

That's due to 0.25 = 1/4 so you just need to rewrite 0.25 as 1/4 to simplify the process and get the right answer

Don't really understand your explanation?

[Springyboy]

Okay since can be rewritten as
Then can be rewritten as
Also, since then can be rewritten as

From there you can just take the integral, therefore which is the answer the textbook gives