Maths Methods 3/4 Help Thread 2011

[onur369]

thanks xZero, one more question while im still doing this exercise

Find x if;
8e^-x - e^x = 2

Make the the indices all postive and than sub in y or what ever value u want and then solve for it. After you find values make them equal to e^x and than find x.

[brightsky]

8e^(-x) - e^x = 2
8/e^x - e^x = 2
8/e^x - e^2x/e^x = 2
(8 - e^(2x))/e^x = 2
8 - e^(2x) = 2e^x
e^(2x) + 2e^x - 8 = 0
To make it easier to understand let's make e^x = u
So the equation becomes:
u^2 + 2u - 8 = 0
(u-2)(u+4) = 0
u = 2 or u = -4
e^x = 2 or e^x = -4
We know that e^x > 0
So ditch the negative solution
So we are left with e^x = 2
which means x = log_e(2)

[eeps]

Question...

Given that the function has rule of the form and and , find the values of and .

Any help would be appreciated!

[Aurelian]

Question...

Given that the function has rule of the form and and , find the values of and .

Any help would be appreciated!

Set up two simultaneous equations and solve;

1.        8 = a(1)^2 + b(1)    ---->     8 = a + b
2.        4 = 2a(1) + b          ---->     4 = 2a + b

1 - 2;
4 = -a  ==> a = -4

Sub in;
b = 12

[Milkshake]

Question...

Given that the function has rule of the form and and , find the values of and .

Any help would be appreciated!

Set up two simultaneous equations and solve;

1.        8 = a(1)^2 + b(1)    ---->     8 = a + b
2.        4 = 2a(1) + b          ---->     4 = 2a + b

1 - 2;
4 = a

Sub in;
b = 4

You mean, a=-4, and therefore b=12?

[Aurelian]

Question...

Given that the function has rule of the form and and , find the values of and .

Any help would be appreciated!

Set up two simultaneous equations and solve;

1.        8 = a(1)^2 + b(1)    ---->     8 = a + b
2.        4 = 2a(1) + b          ---->     4 = 2a + b

1 - 2;
4 = a

Sub in;
b = 4

You mean, a=-4, and therefore b=12?

Oops, yes, I do =) Should've checked over that working... thanks!

[Rairiko]

Got a couple of quick question

-Find the rule of the image when y = 1/x^2 + 2 is translated 5 units in the negative direction of the x-axis and 1 unit in the positive direction of the y-axis then dilated by a factor of 2 from the y-axis.

-Find the rule of the image when y = | x – 1 | + 4 is reflected in the y-axis, dilated by a factor of 3 parallel to the y-axis and translated 2 units in the positive direction of the x-axis.

My answers were y=1/2(x+5)^2 +3  and |-x/3 +1|+6. Are they correct?

[brightsky]

(x,y) --> (2(x' - 5), y' + 1)
x = 2(x'-5) --> x' = (x+10)/2
y = y'+1 --> y' = y - 1

Sub that in: y - 1 = 1/((x+10)/2)^2 + 2 --> y = 4/(x+10)^2 + 3

(x,y) --> (x'+2, -3y')
x = x'+2 --> x' = x-2
y = -3y' --> y' = -y/3

Subbing that in: -y/3 = |(x-2)-1| + 4 --> y = -3|x-3|-12

[ech_93]

I know this question should be really simple, but I can't find the right answer to it

Simplify y=1+log(10)2-2log(10)5 in terms of log.

thanks!

[kefoo]

Got some questions...

1. cosx = sqrt(3)/2, solve if domain is [-pi, pi]
2. For pi/2 < x < pi with cos x = -cos (pi/6), find values of x.

[Andiio]

I know this question should be really simple, but I can't find the right answer to it

Simplify y=1+log(10)2-2log(10)5 in terms of log.

thanks!

y = 1 + log_10_(2) - 2log_10_(5)
y = log_10_(10) + log_10_(2) - log_10_(25)
y = log_10_(20/25)
y = log_10_(4/5)

[luken93]

Got some questions...

1. cosx = sqrt(3)/2, solve if domain is [-pi, pi]
2. For pi/2 < x < pi with cos x = -cos (pi/6), find values of x.

1.

<-- Not within domain, so minus period (2pi)


2.


<-- Not within domain, so minus period (2pi)

[david10d]

Hey guys,

For the function f: R -> R, f(x)=-(2-x)^2(x+1)(3x-1), find the values of x for which f is positive.

[xZero]

work out the x-intercept by letting f(x)=0, then either graph it to see where f(x) is positive or just think of it logically.

The sign of f(x) will be alternating every time it crosses the x-axis so it goes something like: positive until it reaches the first x-intercept from the left hand side, it will be come negative until it reaches the second x-intercept and so on

[m@tty]

(2-x)^2 is always positive in R, so you can ignore it.

Then you're left with -(x+1)(3x-1)

Because of the minus sign, you need the product of the two brackets to be negative.
Ie.

x+1<0 AND 3x-1>0

OR

x+1>0 AND 3x-1<0

The solution is the union of the above intersections.