Question

How to find the x-coordinates?

Answer 1

Please see below.

Explanation:

.

`y=(x^2-1)^n`, `n >1`

We take the first derivative of the function, set it equal to `0`, and solve for `x`:

`dy/dx=2xn(x^2-1)^(n-1)=0`

`2xn=0, :. x=0`

`x^2-1=0, :. x=+-1`

These are the `x`-coordinates of the function's maximums and minimums.

We take the second derivative of the function, set it equal to `0`, and solve for `x` to find the `x`-coordinates of the inflection points:

`(d^2y)/dx^2=2n(x^2-1)^(n-1)+4x^2n(n-1)(x^2-1)^(n-2)=0`

`2n(x^2-1)^(n-2)(x^2-1+2x^2(n-1))=0`

`2n(x^2-1)^(n-2)(x^2(2n-1)-1)=0`

`2n(x^2-1)^(n-2)=0, :. x=+-1`

`x^2(2n-1)-1=0, :. x=+-sqrt(1/(2n-1))`

for `n=2, 3, 4, 5, 6`

we have `x=+-sqrt(1/3), +-sqrt(1/5), +-sqrt(1/7), +-sqrt(1/9), +-sqrt(1/11)` as `x` coordinates of inflection points of the functions in addition to `+-1`.

You can see all this clearly in the graph of the family of functions for `n=2, 3, 4, 5, 6` as shown below:

`n=2` is the purple curve.

`n=3` is the green curve.

`n=4` is the blue curve.

`n=5` is the red curve.

`n=6` is the pink curve.

Answer 2

`a=2, b=-1, c=-3`

Explanation:

.

Geish,

If we calculate the right hand side of the matrix equation and set it equal to the left hand side which is the transformation matrix we find the following transformation:

`x -> x+b`

`y -> ay+c`

This means in that if we plug these values in the `g(x)` we should end up with `h(x)`:

`g(x)=y=2/(x+1)-3`

`ay+c=2/(x+b+1)-3`

Let's solve for `y`:

`ay=(2-3x-3b-3-cx-cb-c)/(x+b+1)`

`ay=((-3-c)x+2-3b-3-cb-c)/(x+b+1)`

`h(x)=y=(((-3-c)x+2-3b-3-cb-c)/a)/(x+b+1)=1/x`

`(((-3-c)x+2-3b-3-cb-c)/a)/(x+b+1)=1/x`

For the two sides to be equal, we should have the numerators to be equal and the denominators to be equal:

`x+b+1=x, :. b+1=0, :. b=-1`

`((-3-c)x+2-3b-3-cb-c)/a=1`

`-3-c=0, :. c=-3`

`(2-3b-3-cb-c)/a=1`

`(2-3(-1)-3-(-3)(-1)-(-3))/a=1`

`(2+3-3-3+3)/a=1`

`2/a=1`

`a=2`