1. ## function

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

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
Try plotting the function on your calculator.

You can see no value of y is attained more than once, so the function is onto. And every possible value of y is attained, so the function is one to one. I don't quite understand what d) is asking, but a) is true.

3. Originally Posted by robeuler
You can see no value of y is attained more than once, so the function is onto. And every possible value of y is attained, so the function is one to one.
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:

. . . . . $0\, =\, \frac{(2x_2\, +\, \sin{(x_2)})\, -\, (2x_1\, +\, \sin{(x_1)})}{x_2\, -\, x_1}$

Multiply through by the denominator to get:

. . . . . $0\, =\, (2x_2\, +\, \sin{(x_2)})\, -\, (2x_1\, +\, \sin{(x_1)})$

. . . . . $2x_1\, +\, \sin{(x_1)}\, =\, 2x_2\, +\, \sin{(x_2)}$

. . . . . $2x_1\, -\, 2x_2\, =\, \sin{(x_2)}\, -\, \sin{(x_1)}$

. . . . . $2\, =\, \frac{\sin{(x_2)}\, -\, \sin{(x_1)}}{x_1\, -\, x_2}$

But is this value possible?