Prelim 2016 Maths Help Thread (1 Viewer)

[eyeseeyou]

For 1 I used the differentation of the surd rule and got 1/ 2* squareroot of (x-3)

Could you retry the question by subbing it back in bc when I sub it back in it doesn't seem to work

 

[InteGrand]

For 1 I used the differentation of the surd rule and got 1/ 2* squareroot of (x-3)

Could you retry the question by subbing it back in bc when I sub it back in it doesn't seem to work

 

[eyeseeyou]

How do you simplify this further g(x)=x^3

I got yp to 1/3x^-2/3

 

[InteGrand]

How do you simplify this further g(x)=x^3

I got yp to 1/3x^-2/3

 

[eyeseeyou]

Can't you make it into a surd form?

 

[eyeseeyou]

BTW if anyone here has nothing to do please post in the "count to 100 before a moderator posts" thread

 

[leehuan]

Can't you make it into a surd form?

But not necessary

 

[eyeseeyou]

BTW integrand the answer for 1 was 1/2*squareroot of (x-3)

 

[leehuan]

BTW integrand the answer for 1 was 1/2*squareroot of (x-3)

That's what he said.

 

[InteGrand]

BTW integrand the answer for 1 was 1/2*squareroot of (x-3)

I know. I got that (recall I said you could change the y's to x at the end).

 

[eyeseeyou]

I know. I got that (recall I said you could change the y's to x at the end).

Basically all you had to do was find the inverse and sub in x into the equation

 

[leehuan]

Basically all you had to do was find the inverse and sub in x into the equation

You actually don't unless the question specifically asks for g^-1(x), marking g^-1(y) as wrong.

 

[eyeseeyou]

dw I solved it now

For g^-1(x) I got 1/3x^(2/3)

 

[eyeseeyou]

1. f(x)=(x-1)^2+3, x>=1

Find f^-1'(x),x>3

2. g^-1'=x^3

Find g^-1'(x)

3. h(x)=x^3-3x, -1<x<1 find h^-1'(0)

I am getting stuck with this

What I did was y=x^3-3x, -1<x<3
then x=y^3-3y, -1<y<3

Then what do I do Integrand?

 

[InteGrand]

I am getting stuck with this

What I did was y=x^3-3x, -1<x<3
then x=y^3-3y, -1<y<3

Then what do I do Integrand?

Like I said before, it is impractical to try inverting that function.

The answer is just 1/h'(0), because h(0) = 0 (a = 0 and b = 0 in the Inverse Function Theorem formula). Since h'(0) = -3, the answer is -1/3.

 

[eyeseeyou]

Like I said before, it is impractical to try inverting that function.

The answer is just 1/h'(0), because h(0) = 0 (a = 0 and b = 0 in the Inverse Function Theorem formula). Since h'(0) = -3, the answer is -1/3.

what?

 

[InteGrand]

what?

Check the post I wrote where I did your Q1, I wrote it there.

 

[eyeseeyou]

Consider the function f(x)=(x^3+8x)/8
a. Show that f(x) is an increasing function
b. Find the equation of the tangent to y=f(x) at the origin
c. Using equal scales of axes, sketch a graph of y=f(x) over the domain -2=<x=<2 and draw its tangent at the origin
d. explain why an inverse exists and draw a grapg of y=f^-1(x) on your diagram
e. Solve f^-1(x)=8
f. Evaluate Integral of 3 to 0 f^-1(x) dx

 

[eyeseeyou]

BTW Integrand just realised but you've been here all night lol

 

[Trebla]

eyeseeyou you still haven't answered my questions