Solving equations involving e? (1 Viewer)

[hscishard]

This is from Year 12 Cambridge 2unit Chp 2F Q16.

Show that the curves y=e^x and y = (e-1)x + 1 meet at (0,1) and (1,e)

 

[cutemouse]

Just sub the points into the equations.

 

[hscishard]

Are you sure we're allowed to do that in "Show" questions.

 

[bleaver]

Showing that those equations have identical y values for particular x values (ie subbing into the equations separately) and visa versa is fine and accepted.
The other way would be to equate your two equations...goodluck with that haha
So yeah, that would be fine.

 

[waryap]

This is from Year 12 Cambridge 2unit Chp 2F Q16.

Show that the curves y=e^x and y = (e-1)x + 1 meet at (0,1) and (1,e)

use the substitution method
y=e^x ---------1
y=(e-1)x + 1 --------2
sub 1 into 2
e^x = ex - x + 1
e^x -ex + x -1 = 0
sub the values in and the equation should always = 0
when x = 0
e^0 -e(0) + o -1 = 0
1 - 1 = 0
when x =1
e - e + 1 - 1 = 0

not sure if I'm correct in doing it this way hahaha... correct me if I'm wrong, please. Although I do think I'm doing this wrongly.

 

[bleaver]

use the substitution method
y=e^x ---------1
y=(e-1)x + 1 --------2
sub 1 into 2
e^x = ex - x + 1
e^x -ex + x -1 = 0
sub the values in and the equation should always = 0
when x = 0
e^0 -e(0) + o -1 = 0
1 - 1 = 0
when x =1
e - e + 1 - 1 = 0

not sure if I'm correct in doing it this way hahaha... correct me if I'm wrong, please. Although I do think I'm doing this wrongly.

That method is fine, except you'd have to put a statement at the end stating that as LHS=RHS when x=0 and x=1, then the equations meet at these x values.
Your method is basically a different way of doing the same thing.

HOWEVER! - Considering that you've eliminated the 'y' variable from the equations when you equated, you need to go back and substitute back into one of original equations to demonstrate that the y value is 1 when x=0 and e when x=1. Just to be 'technically' correct.
Otherwise there is nothing wrong with your approach.