VCE Methods Question Thread!

[Phenomenol]

i tried giving it a shot but got upto the second line with -x^2+5x...

Something like this? (pardon the large image)

[lzxnl]

y=3 cos 2 ( x + pie over 2) - 1
 

* what software are you guys using to show your mathematical notation?

Dear me. I don't know how to take the cosine of food, but if you ever work that out do tell me



So type in y=3\cos{2(x+\frac{\pi}{2}) - 1

put [tex] on one side and [ /tex] on the other side of the expression (without the gap between the bracket and the dash. leaving it out would make my font appear funny)

[Einstein]

Something like this? (pardon the large image) (Image removed from quote.)

yeah thats correct! never knew you could have it as fractions , i knew that step, but haven't come across a question that required a fraction therefore i was second guessing haha 

[Frozone]

Dear me. I don't know how to take the cosine of food, but if you ever work that out do tell me



So type in y=3\cos{2(x+\frac{\pi}{2}) - 1

put [tex] on one side and [ /tex] on the other side of the expression (without the gap between the bracket and the dash. leaving it out would make my font appear funny)

Oh wow how embarassing!
Now could you please help me graph the equation. The translation of the graph is extremely confusing.

[lzxnl]

Oh wow how embarassing!
Now could you please help me graph the equation. The translation of the graph is extremely confusing.

Your function is y = 3 cos(2x + pi) - 1
Using symmetry properties of cosine, this is y = -3 cos (2x) - 1
Is that any easier for you to sketch?

[Spiritual]

1) The function with rule f(x) √2x-k has implied (maximal) domain: _?

2) The function with rule f(x) = mx + 2, m < 0, has an inverse function with rule f-1(x) = ax + b, a, b, is an element of R. Which one of the following statements is true?
A. a > 0, b > 0
B. a < 0 , b < 0
C. a > 0, b < 0
D. a < 0, b > 0
E. a = 1/m^2, b = -2

Thanks in advance!!

[Orb]

1) The function with rule f(x) √2x-k has implied (maximal) domain: _?

2) The function with rule f(x) = mx + 2, m < 0, has an inverse function with rule f-1(x) = ax + b, a, b, is an element of R. Which one of the following statements is true?
A. a > 0, b > 0
B. a < 0 , b < 0
C. a > 0, b < 0
D. a < 0, b > 0
E. a = 1/m^2, b = -2

Thanks in advance!!

1) As it's root 2x-k, a square root cannot be negative for real numbers, so we want to find the values of x that make the root positive or 0.

So set up the equation 2x-k greater or equal to 0.

Solve it to find x greater of equal to k/2
Therefore, maximal domain is [k/2, infinite)

2) The inverse function f-1(x) is also x/m - 2/m
Because x/m = 1/m multiplied by x
a= 1/m

because m < 0, 1/m is < 0, a < 0
because b= -(2/m)
therefore, b > 0 (negative of a negative number is positive)

Hence, question's answer is D

[e^1]

how would i solve the following question for:
0<f(|x|)<k ?

Hopefully I have corrected this solution for future viewers, even though it is long overdue.







For .

However, we note that if , . When , also. So is true for all .


For :



Applying log on both sides of an inequality is explained here.

So the possible solutions for x are where , for the inequality to be true.
(If you prefer, )

[soNasty]

thanks for the answer e^1

also, can someone help me with solving sin(x/2)=-1/2  over x ~ [-2pi,2pi]
i would usually do it this way:

solving for the basic angle to get BA=pi/6      - change domain to -pi<=x<=pi
note that sin is negative, so we'd get 7pi/6 and 11pi/6, however they dont fit within the noted domain.
theres another, correct way to solve this question and it involves using -pi and -pi+pi/6.. which i dont seem to get.. (not sure if ive explained it correctly)

[Phenomenol]

thanks for the answer e^1

also, can someone help me with solving sin(x/2)=-1/2  over x ~ [-2pi,2pi]
i would usually do it this way:

solving for the basic angle to get BA=pi/6      - change domain to -pi<=x<=pi
note that sin is negative, so we'd get 7pi/6 and 11pi/6, however they dont fit within the noted domain.
theres another, correct way to solve this question and it involves using -pi and -pi+pi/6.. which i dont seem to get.. (not sure if ive explained it correctly)

Don't "change the domain". The domain should be the same. so if -2pi<=x<=2pi, then -pi<=x/2<=pi, which is much easier to interpret.
So you need to find values in between -pi and pi, for which sin(value) = -1/2. Let these values equal x/2.

(For sin(y) = -1/2, y = 7pi/6 + 2npi, 11pi/6 + 2npi, where n is any integer (2 solutions per every revolution around the unit circle). For a restricted domain of y, n will be reduced to certain values.)

So, x/2 = -5pi/6, -pi/6

x = -5pi/3, -pi/3

[Spiritual]

1) As it's root 2x-k, a square root cannot be negative for real numbers, so we want to find the values of x that make the root positive or 0.

So set up the equation 2x-k greater or equal to 0.

Solve it to find x greater of equal to k/2
Therefore, maximal domain is [k/2, infinite)

2) The inverse function f-1(x) is also x/m - 2/m
Because x/m = 1/m multiplied by x
a= 1/m

because m < 0, 1/m is < 0, a < 0
because b= -(2/m)
therefore, b > 0 (negative of a negative number is positive)

Hence, question's answer is D

Thank you very much.

[kx4y]

When we're sketching two graphs on one set of axes (eg. 1 exponential and 1 log graph), how do we know if there are intersections or not?

[alchemy]

When we're sketching two graphs on one set of axes (eg. 1 exponential and 1 log graph), how do we know if there are intersections or not?

Equate the equation of the first graph to the equation of the other. If a solution exists, that's the point of intersection.

[jessss0407]

Hey guys! I need help with this question:
1a) For the functions f and g that are defined and differentiable for all real numbers, it is
known that:
f(1)=6, g(1)=−1, g(6)=7 and f(−1)=8
f'(1)=6, g'(1)=−2, f'(−1)=2 and g'(6)=−1
Find:
i) (f◦g)'(1)
ii) (g◦f)'(1)
iii) (fg)'(1)
iv) (gf)'(1)
v) (f/g)'(1)
vi) (g/f)'(1)

Thanks!

[Phenomenol]

Hey guys! I need help with this question:
1a) For the functions f and g that are defined and differentiable for all real numbers, it is
known that:
f(1)=6, g(1)=−1, g(6)=7 and f(−1)=8
f'(1)=6, g'(1)=−2, f'(−1)=2 and g'(6)=−1
Find:
i) (f◦g)'(1)
ii) (g◦f)'(1)
iii) (fg)'(1)
iv) (gf)'(1)
v) (f/g)'(1)
vi) (g/f)'(1)

Thanks!

i) and ii) involve the chain rule.
iii) and iv) involve the product rule.
v) and vi) involve the quotient rule.
Write out the general forms and then substitute correctly.

If you need more help then I most certainly can.