VCE Specialist 3/4 Question Thread!

[alchemy]

Find dy/dx of 2sqrt(x)sin^-1(x^2), giving your answer in the simplest form.
My working out is attached.
The answer is ((4x^3/2)/sqrt(1-x^4)) + (sin^-1(x^2)/sqrt(x))

[kinslayer]

Find dy/dx of 2sqrt(x)sin^-1(x^2), giving your answer in the simplest form.
My working out is attached.
The answer is ((4x^3/2)/sqrt(1-x^4)) + (sin^-1(x^2)/sqrt(x))

You forgot to put du/dx out the front of the derivative of the inverse sine.

edit: couldn't get the LaTeX to show up in post, attached

[Eugenet17]

how do i find the maximal domain of cos^-1(6/x) algebraically?

[alchemy]

how do i find the maximal domain of cos^-1(6/x) algebraically?

The domain of cos^-1(x) is [-1,1]
So in order for cos^-1(6/x) to exist: -1<=6/x<=1.
Solve this to get the maximal domain.

[Eugenet17]

I knew that but how do i solve from there?

[Eugenet17]

Answer is x<=-6 and x>=6 D:

[kinslayer]

I knew that but how do i solve from there?

You have two inequalities: 1) -1 <= 6/x and 2) 1 >= 6/x. You can work with these just like normal equations except if you multiply both sides by a negative number then the inequality will flip.

When x > 0, 1) tells us that -x <= 6 (not useful) and 2) tells us that x >= 6.

When x < 0, 1) tells us that -x >= 6, while 2) tells us that x <= 6 (not useful).

Now when x > 0, the absolute value |x| = x, but if x < 0, |x| = -x. So putting it all together the solution is {x: |x| >= 6}. It would also be correct to put {x: x <= -6 or* x >= 6} but I think it's cleaner with the absolute value sign there.

*-edited

[Eugenet17]

You have two inequalities: 1) -1 <= 6/x and 2) 1 >= 6/x. You can work with these just like normal equations except if you multiply both sides by a negative number then the inequality will flip.

When x > 0, 1) tells us that -x <= 6 (not useful) and 2) tells us that x >= 6.

When x < 0, 1) tells us that -x >= 6, while 2) tells us that x <= 6 (not useful).

Now when x > 0, the absolute value |x| = x, but if x < 0, |x| = -x. So putting it all together the solution is {x: |x| >= 6}. It would also be correct to put {x: x <= -6 and x >= 6} but I think it's cleaner with the absolute value sign there.

Thanks so much!

[rhinwarr]

Can someone explain the answer to this question please?

[brightsky]

the function y = arctan(x) has a range of (-pi/2, pi/2). this means that the 'formula' Arg(a+bi) = arctan(b/a) only works if the complex number a+bi is in the first or fourth quadrant, i.e. has an Argument which lies within the range (-pi/2,pi/2). hence, a must be positive but b can be anything.

[Eugenet17]

f(x)= 2x^3+6x^2-12, use f''(x) to find the coordinates where the gradient is a minimum (the point of inflexion).

How do I do this and why is the point of inflexion where the gradient is a minimum?

[EspoirTron]

f(x)= 2x^3+6x^2-12, use f''(x) to find the coordinates where the gradient is a minimum (the point of inflexion).

How do I do this and why is the point of inflexion where the gradient is a minimum?

Firstly you find f'(x), which is equal to 6x^2+12x. Then we find f''(x), which is equal to 12x+12. To find the point of inflection we solve that f''(x)=0 which gives us that x=-1. Now, let's use a sign diagram. Let's let x=-2: f''(-2) = -12, which is less than 0, so around this region the graph is concave down. Let us now allow x=0: f''(0)=12, which is greater than 0, and is therefore concave up. Since we change concavity about x=-1, we know that this is a point of inflection on the graph of f(x).

To your second point. This is not a stationary point of inflection, and thus f'(x) does not equal 0 at x=-1. However, the gradient is at a minimum. How? Well that is seen through the change of concavity. At this point someone can correct me if I am wrong. However, as we approach x=-1 our gradient starts becoming smaller and smaller (its magnitude), but it never equals to zero. At x=-1 the graph 'kinks' and goes to concave up. Around this local region the magnitude of the gradient assumes very small values. We can go as small as we would like, but since the point isn't stationary it won't equal to zero. Think about drawing an infinite set of tangents around this point. What you will find is the slope - or, more precisely the gradient - will keep getting smaller and smaller, but it will never be horizontal. This point proves that the gradient keeps getting smaller but never becomes zero.

I hope that helped

[Eugenet17]

Firstly you find f'(x), which is equal to 6x^2+12x. Then we find f''(x), which is equal to 12x+12. To find the point of inflection we solve that f''(x)=0 which gives us that x=-1. Now, let's use a sign diagram. Let's let x=-2: f''(-2) = -12, which is less than 0, so around this region the graph is concave down. Let us now allow x=0: f''(0)=12, which is greater than 0, and is therefore concave up. Since we change concavity about x=-1, we know that this is a point of inflection on the graph of f(x).

To your second point. This is not a stationary point of inflection, and thus f'(x) does not equal 0 at x=-1. However, the gradient is at a minimum. How? Well that is seen through the change of concavity. At this point someone can correct me if I am wrong. However, as we approach x=-1 our gradient starts becoming smaller and smaller (its magnitude), but it never equals to zero. At x=-1 the graph 'kinks' and goes to concave up. Around this local region the magnitude of the gradient assumes very small values. We can go as small as we would like, but since the point isn't stationary it won't equal to zero. Think about drawing an infinite set of tangents around this point. What you will find is the slope - or, more precisely the gradient - will keep getting smaller and smaller, but it will never be horizontal. This point proves that the gradient keeps getting smaller but never becomes zero.

I hope that helped

Thanks for the clarification, makes alot more sense now

[Eugenet17]

Also, just to clarify, if it's a stationary point of inflection there is still a change of concavity except the gradient will =0?

[EspoirTron]

Also, just to clarify, if it's a stationary point of inflection there is still a change of concavity except the gradient will =0?

I am glad that you understand it now
Yes, even if it is a stationary point of inflection there will still be a change in concavity. It could be from concave up to concave down, or vice versa. If there isn't a change in concavity it won't be an inflection point. Local maxima and minima don't have changes in concavity about f'(x)=0.