Mr. Study's 3 and 4 Questions

[Mr. Study]

oO, You obliterated those questions easily... Thanks for the help Rohit
No one solve 3. I THINK I know what to do now. xD, NVM.

[brightsky]

e^x - 4e^(-x) = 0
e^x - 4/e^(x) = 0
e^(2x) - 4 = 0 (note that e^x cannot equal 0)
e^(2x) = 4
x = 1/2 * ln(4)

[Mr. Study]

Quartic's Question

1. . Is there any way to find the turning points WITHOUT a CAS calculator?
                                                                                                           _
2. Rewrite as hybrid. I get this far:
                                                                                                          |_.
I'm not too sure how to write the less/greater than for x.

Thanks.

[monkeywantsabanana]

Quartic's Question

1. . Is there any way to find the turning points WITHOUT a CAS calculator?

you derive it then equate it to 0, solve for x, sub it back into original equation to find the y.

[Panicmode]

Quartic's Question

1. . Is there any way to find the turning points WITHOUT a CAS calculator?
                                                                                                           _
2. Rewrite as hybrid. I get this far:
                                                                                                          |_.
I'm not too sure how to write the less/greater than for x.

Thanks.

2. Think about it in terms of where it "cuts" the negative region. Since is a positive quadratic, it will be positive for x < x-intercept 1 and x> x intercept 2. It will be negative between those values.





so it will be for and

and
for

[b^3]

2. Rewrite as hybrid. I get this far:
                                                                                                          |_.
I'm not too sure how to write the less/greater than for x.

Thanks.

Now remember a mod makes whatever is inside it positive, so its is negative (y<0) then it will be made positive.

So when then you are just left with the positive of it, i.e. . When , it i s negative and being made positive hence we have .

So you have to work out when
Because it is not linear, we need to draw it out and visually work out when it is above 0. So intercepts are needed x(x-3)=0, x=0,3
https://www.desmos.com/calculator/bntqyrvf32

So it is above or equal to 0 when x<=0 and x=>3
So that is your domain for the
For it will be when its less then 0, so 0<x<3

Hope that helps

EDIT: Beaten.

[Mr. Study]

. Gradient is 5 at (2,-2). Find a and b

This is what I have done:



Let


Let y=-2 and x=2 and sub into first equation.


I am not too sure what to do next. Also, If My working out is either wrong or I'm reasoning the wrong way, please say so.

Thank you.

EDIT: Just need my working out to be checked.

Find acute angle lying between and











I let m₁=-3/4 and m₂=2/3.


But I know we can't get tan^-1(negative number). So would someone be able to look through what I've done so far?

Thanks.

[illuminati]

. Gradient is 5 at (2,-2). Find a and b

This is what I have done:



Let


Let y=-2 and x=2 and sub into first equation.


I am not too sure what to do next. Also, If My working out is either wrong or I'm reasoning the wrong way, please say so.

Thank you.

EDIT: Just need my working out to be checked.

Okay so you have
dy/dx = 5
and then you go on to
5 = 2ax - bx^(-2)
you can sub in x = 2 here, because the gradient at x = 2 is 5
so you'll get
5 = 4a - b/4 ----> 20 = 16a - b
and then you have two simultaneous equations with a and b.

[illuminati]

Find acute angle lying between and











I let m₁=-3/4 and m₂=2/3.


But I know we can't get tan^-1(negative number). So would someone be able to look through what I've done so far?

Thanks.

You can tan inverse a negative.
You'll get ~-0.64 radians
Your angle is tan^-1(2/3) - tan^-1(-3/4), because if you recall, the angle between a line and the x axis is tan(theta) = m
So you can firstly find the angle your positive gradient line makes with the x axis, and then subtract the negative angle your negative gradient line makes
and you'll get the answer
Hope this helps

[Mr. Study]

Hey,

Thank you so much for that.

I had to change my calculator settings to do the tan inverse, otherwise it was a syntax error. :O