Math Help - point of inflextion,

1. point of inflextion,

A) show that $y = 6sin2x + 4cos2x$ satisfies the equation
$\frac{d^{2}y}{dx^{2}} +4y = 0$

b)the expression $y = 6sin2x + 4cos2x$ can be written as $R sin(2x + a)$ where R and α are positive constants,
0 < α < pi/2. Find the values of R and α, correct to 3 decimal places.

(c) What is the smallest positive value of x where y has a point of inflection?
I have done part a and b, but need help with part c.

$\sqrt{52}sin(2x + 0.588)$

$\frac{d^{2}y}{dx^{2}} = -24sin2x - 16cos2x$

Thank you.

2. The seond derivative is zero if $\tan(2x)=\frac{-2}{3}$. WHY?

3. Originally Posted by Plato
The seond derivative is zero if $\tan(2x)=\frac{-2}{3}$. WHY?
How did you get tan2x?

I got $tan \alpha = \frac{2}{3}$

Thanks.

4. Originally Posted by Tweety
How did you get tan2x?
$- 24\sin(2x) - 16\cos(2x)=0$
$- 24\sin(2x) = 16\cos(2x)$
$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}= \frac{16}{-24}$

5. Originally Posted by Plato
$- 24\sin(2x) - 16\cos(2x)=0$
$- 24\sin(2x) = 16\cos(2x)$
$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}= \frac{16}{-24}$
Okay thanks, I am not quite sure why this result would make the second derivative zero?

I know how to find the coordinates of a stationary point, but don't get what the question is asking, is it asking me to find all the points of inflection on the curve
$y = 6sin2x + 4cos2x$ and than find the lowest value of x corresponding to a point of inflection?

thank you for clarifying.

6. Originally Posted by Tweety
I am not quite sure why this result would make the second derivative zero?
That is how we find points of inflections.

7. Originally Posted by Plato
That is how we find points of inflections.
Yes I know this, but how does tan2x = -2/3 make the second derivative equal zero? And how would I work out the lowest value of x from this?

thanks,

8. Originally Posted by Tweety
how would I work out the lowest value of x from this?thanks,
$x=\frac{1}{2}\left(\arctan\left(\frac{-2}{3}\right)+\pi\right)$