VCE Methods Question Thread!

[ETTH96]

Hey guys, can someone help me find the inverse of this function: f(x)=  8/(2x+1)^2         (x > -1/2)

[lzxnl]

Hey guys, can someone help me find the inverse of this function: f(x)=  8/(2x+1)^2         (x > -1/2)

y=8*(2x+1)^-2
Swap x and y
x=8*(2y+1)^-2
(x/8)^(1/2)=2y+1 (take positive square root because the original function has x>-1/2 => inverse has y>-1/2)
y=1/2*((x/8)^1/2-1)
Simply as much as you want after.

Hey guys, can someone help me out with these transformations of a truncus

From y=1/x^2  -->   y=8/(2x+1)^2  -1

Is this right?
- dilation factor 8 from x axis
- dilation factor 1/2 from y axis
- translation 1/2 units left
- translation 1 unit down

I'm a bit confused about dilation from the y axis.. Can someone clarify how to find this? Thanks!

The translations are correct. Dilation from x axis means the y values are affected and dilation from y axis means the x values are affected.
Using the substitution method, if you replace x with x/a, you dilate from the y axis by factor a. And vice versa for dilating from the x axis.

[LOLs99]

Hey guys, can someone help me find the inverse of this function: f(x)=  8/(2x+1)^2         (x > -1/2)

Let y= f(x)
Inverse: swap x & y.

Spoiler

 






Sketch the original graph and then observe whether to reject positive answer or negative answer.
Since x>-1/2 is above the y-axis therefore reject -ve answer.

[ETTH96]

Thanks guys, really helpful!

How would I change the function 3x(x-2)^2  into the form of (ax+b)(x-2)^2 ?? Thanks

so basically, what is the value of a and b equal to? im so lost.. is it just a=3 and b=0?

[lzxnl]

There's a question asking you that!?
I would have responded the way you did too.

[Jason12]

Is (x+ 1)x^2 the same as x^2(x+1)?

[alchemy]

Is (x+ 1)x^2 the same as x^2(x+1)?

Yeah. It's like 2(3) and 3(2). They are both the same thing. In this case expanding either one will yield x^3 + x^2.

[Stevensmay]

Hey guys, how do I go about doing this question?

f(x)=x(x-2)^3  +2
The coordinates of the stationary points of the graph of y=f(x) are (u,2) and (v, 5/16). find the values of u and v

are stationary points turning points?
If so, would I differentiate f(x) and then sub in the coordinates given and solve simultaneously..? Please help!

Stationary points can be turning points. They can also be points of inflection but I don't think that is covered in methods?

In this question you would differentiate f(x).



Seeing as we are finding the stationary points, we know the gradient at these points will be zero.
Let f'(x) = 0, which will give us the x values when the gradient of this function is zero.



Solve this and we find that

All we need to do now is match these x values to the correct y value, so we know which co-ordinate it goes with.
Simply substitute in one of the values and you will get one of the y co-ordinates given.

[ETTH96]

Thank you stevensmay, i managed to figure it out in the end! thanks anyway for reassuring me!

Can someone help me state the transformations of a graph to another graph:

x(x-2)^3 +2   ---->     2x(2x-2)^3

I know theres a translation of 2 units down
What about the dilations?
is it 2 from the y axis and 1/2 from the x?  I'm confused because theres two x's...

[nhmn0301]

Thank you stevensmay, i managed to figure it out in the end! thanks anyway for reassuring me!

Can someone help me state the transformations of a graph to another graph:

x(x-2)^3 +2   ---->     2x(2x-2)^3

I know theres a translation of 2 units down
What about the dilations?
is it 2 from the y axis and 1/2 from the x?  I'm confused because theres two x's...

As you can see from the inititial equation:
X has been replaced with 2x. Hence, there is a dilation of factor 1/2from the y-axis.
Next, 2x-2 can be written in the transformation form as 2(x-1). Hence, this assures me that there is definitely a dilation of factor 1/2 from the y-axis. You can also identify that there is a translation of 1 unit to the  negative direction in the x-axis from the ORIGINAL GRAPH and 2 units to the negative direction of the y-axis from the ORIGINAL GRAPH.
Hope this helps.

[Bluegirl]

Could someone please give me a hint?

Solve for x in each of the following, giving exact answers:

ln(2x-3) = 0

Not sure what to do.

[Stick]

[b^3]

HINT: Change it from a log into an exponential, "log base of the answer gives the power".

Spoiler

EDIT: Beaten.

[Bluegirl]

Thanks!

Now for
ln(3x-2) = ln(x=1)

Was I right in making it ln(3x-2/x=1)
What do I do now?

[rhinwarr]

I don't know how you can have x=1 in the ln but anyway... I'll assume you meant minus

Ln(3x-2/x-1) = 0
3x-2 / x-1 = e^0
3x-2 / x-1 = 1
3x-2 = x-1
2x = 1
x = 1/2