CPhill

x^3-x^2+x+1 =x^3+x^2+x-1 

Subtract x^3, x   from each side

-x^2 + x + 1  =  x^2 - 1

-x^2 + 1  = x^2 - 1    rearange as

0  =  2x^2 - 2

2x^2 - 2  = 0     divide through by 2

x^2 - 1  = 0    factor

y = (1)^3 + (1)^2 + 1 - 1  =

(x + 1) ( x - 1)  = 0 

Set each factor to 0 and solve for x

And  x = 1   or x = -1

And using  x^3+x^2+x-1   

y = (1)^3 + (1)^2 + 1 - 1  =  2

y = (-1)^3 + (-1)^2 - 1 - 1  =  -2

So the solutions are  (1, 2)  and (-1. -2)

f^-1 (4)  =  3

h(7)  =  5

So

h (h (7) )  =  h (5)  =   - 1

(x^2 + 2x - 1)(x^2 + 2x + 5) =

x^2 * x^2    + x^2 *2x +  5x^2   =  x^4 + 2x^3 + 5x^2    (1)

2x*x^2 + 2x*2x + 10x  =  2x^3 + 4x^2 + 10x   (2)

-1*x^2 - 1*2x - 1*5  =   -x^2 - 2x - 5   (3)

Combining (1) (2) (3)  we have

x^4 + 4x^3  + 8x^2 + 8x - 5

The slope - except at (0,0) -  is always positive so there are no  "extremes"

The inflection point (0,0) is not an "extreme"......the slope would have to change on either side of this point, but it doesn't.....the inflection point just represents the point where - in this  case - the curve changes from concave down to concave up

 Rotate 90 degrees counterclockwise about the orgin.

Translate 4 units left.

Thanks, X².....that notation often confuses me, too...!!!!

Your  pic and presentation of the answer on this geometry question made it pretty easy to see......most of these proofs rely one one key thing....that exterior angle-opposite interior angle thing was the key...

I saw that one that rosala posted......and your answer...I think that one is probably above my puny level....>>>LOL!!!!!

Yes....I'm never sure are about "equal"  vs. "congruent, either.....

WDQA

AAS Congruence Theorem