Graphs Help (1 Viewer)

[4025808]

May anyone help me with these questions? I'm not sure what they mean and like I'm not sure how to work them out

ii) of the first question (not sure how to get the set values of the real number k such as that f(x) = kx has three real roots)

in the 2nd image, I dont understand ii) of a)
and I don't get how to do b) since i havent done inverse trig functions yet
[find the domain and range of the function tan^-1(e^x), sketch the curve....]

help is greatly appreciated =D

(this is the 2nd image, damn its a bit screwed...)

 

[ohexploitable]

For the first question, 0 < k < 1

 

[ohexploitable]

It just means which straight lines passing through the origin will intersect the curvy line three times.

 

[xV1P3R]

I think ohexploitable means 0< k <1.

For the 2nd image, part ii)
For part i), you get a max and min turning point. To get 3 real roots, the y value of the max has to be above or on the x axis, and the y value of the min has to be below or on the x-axis. Then you look at the other keyword "different" which means you can't have double or triple roots ie. the y values of the max and min =/= 0.

To do b), it's best if you know the inverse trig functions.

 

[ohexploitable]

I'm not sure on the proper method because I haven't done this yet.

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You could sketch x³+6x²+9,

Max: (-4,41)
Min: (0,9)

Adding a constant 'k' would move it up or down, you want to move it so it intersects the x-axis 3 times with no double roots. So, -41 < k < -9.

 

[ohexploitable]

I think ohexploitable means 0< k <1.

Wut?

 

[hscishard]

domain x any real no.
range 0 < y< pi/2

Range of tan(-1)x is -pi/2< angke < pi/2

But because e^x is always positive, tan^-1e^x can never be negative. Hence the range is between pi/2 and 0.
As e^x approaches infinity, it approaches pi/2
As e^x approaches to 0 (when x approaches -infinity), it approaches 0

The derivative of tan(-1) u is 1/1+u^2 x du/dx
there are no stationary points
When x=0, y= pi/4
You will now have enough info if you've done this inverse stuff.

 

[hscishard]

sin2y/2 = sinycosy

y=tan-1e^x

sin(tan^-1_e^x) x cos(tan^-1_e^x)

Using a rt < triangle, you'll find the hypotentuse is (1+e^2x)^0.5 you should be able to do it from there

 

[hscishard]

I think for a ii) it's 0 < k < 4

 

[hscishard]

I'm not sure on the proper method because I haven't done this yet.

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You could sketch x³+6x²+9,

Max: (-4,41)
Min: (0,9)

Adding a constant 'k' would move it up or down, you want to move it so it intersects the x-axis 3 times with no double roots. So, -41 < k < -9.

It's 9x.
Lol gotta hate it when that happens

Wait this is four unit?? Feels more prelim 2u and hsc 3u