let function f:R to R be defined by f(x)=2x+sinx for real x. then f is
a) one to one and onto
b) one to one but not onto
c)onto but not onr to one
d) neither of the common tangent to the curves
I think this may be a bit backwards...? Doesn't "onto" mean "every y-value is 'covered'"...? Doesn't "one-to-one" mean "every y-value corresponds to only one x-value"...?
The conclusion remains the same, however.
Note to original poster: To show, using pre-calculus techniques, that no y-value is covered twice, you might be expected to try to find x-values having the same y-value, and then show that this is impossible. You might be expected to start with the slope formula, noting that the slope between these two (non-existant) points must be zero. Then:
. . . . .$\displaystyle 0\, =\, \frac{(2x_2\, +\, \sin{(x_2)})\, -\, (2x_1\, +\, \sin{(x_1)})}{x_2\, -\, x_1}$
Multiply through by the denominator to get:
. . . . .$\displaystyle 0\, =\, (2x_2\, +\, \sin{(x_2)})\, -\, (2x_1\, +\, \sin{(x_1)})$
. . . . .$\displaystyle 2x_1\, +\, \sin{(x_1)}\, =\, 2x_2\, +\, \sin{(x_2)}$
. . . . .$\displaystyle 2x_1\, -\, 2x_2\, =\, \sin{(x_2)}\, -\, \sin{(x_1)}$
. . . . .$\displaystyle 2\, =\, \frac{\sin{(x_2)}\, -\, \sin{(x_1)}}{x_1\, -\, x_2}$
But is this value possible?